


//* 



GRAPHICAL CALCULUS 



Graphical Calculus 



ARTHUR H. BARKER, BA„ B.Sc. 

SENIOR WHITWORTH SCHOLAR 1895 

LATE ASSISTANT TO THE PROFESSOR OF ENGINEERING, YORKSHIRE COLLEGE, LEEDS 

AUTHOR OF " GRAPHIC METHODS OF ENGINE DESIGN," " THE MANAGEMENT OF 

ENGINEERING WORKSHOPS," ETC. 



WITH AN INTRODUCTION BY 

JOHN GOODMAN, A.M.I.C.E. 

PROFESSOR OF ENGINEERING AT THE YORKSHIRE COLLEGE VICTORIA UNIVERSITY 



SECOND EDITION 



LONGMANS, GREEN, AND CO. 

39 PATERNOSTER ROW, LONDON 

NEW YORK AND BOMBAY 

1902 

All rights reserved 



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INTRODUCTION 

All "up-to-date" teachers of engineering and applied 
sciences generally now recognize the vast superiority 
of graphical over purely mathematical methods of im- 
parting instruction of almost every description. The 
former are much more convincing to the student, 
because they appeal to the eye, the training of which 
is one of the chief objects to be aimed at in the 
education of an engineer. There is no doubt that this 
method is capable of great extension with advantage. 
In this little book, for instance, we see graphical con- 
structions of a very simple character employed to teach 
what, to the beginner, are somewhat abstruse mathe- 
matical principles. 

The attempt to employ purely mathematical, in 
preference to graphical methods, seems to me quite as 
absurd as attempting to teach geography by giving the 
position of towns in terms of their latitude and longi- 
tude, and explaining the shape of a country by giving 
the equation to the coast-line instead of by employing the 
graphical method, i.e. exhibiting a map. The teacher 



vi Introduction. 

who attempted the former method would indeed be 
considered unpractical, and would, I fear, meet with 
but a very limited share of success ; yet, strange to 
say, such a method is precisely that which teachers of 
mathematics are trying to employ with a much more 
subtle subject than geography — the Calculus. Is it, 
then, to be wondered at that many technical students 
shudder at the bare sign of integration, "the long S," 
as they are wont to call it? Not because they cannot 
manipulate the symbols — far from it — but because they 
have not the faintest notion of the physical meaning of 
the processes. 

I have frequently had students come under my 
notice who, although fairly good mathematicians as far 
as bookwork is concerned, yet, through not having had 
the advantage of & practical mathematical training such 
as we give at the Yorkshire College, were utterly at sea 
when they came to apply their mathematics to such 
a simple engineering problem as, for instance, finding 
the quantity of water flowing over a V notch ! It is 
primarily to help such to acquire an intelligent working 
knowledge of the Calculus that Mr. Barker has written 
this little book. Even if it have only a tithe of the 
success that its author has had in teaching mathematics 
by this method, it will still be eminently successful. I 
can unreservedly say that this is exactly the style of 
book that I have been wanting to see for years, and 
I believe it will prove to be of very real value to those 



Introduction, vii 

students of engineering who wish to get a stage beyond 
the barest elements of the subject. 

Some of those who know my propensity to scoff at 
the mathematics so commonly drummed into technical 
students, may, on seeing this introduction, exclaim, " Is 
Saul also among the prophets ? " To such I would 
say that if as a student it had fallen to my lot to be 
under such a teacher as Mr. Barker, I should always 
have been numbered among the prophets, though 
possibly the minor ones. 

JOHN GOODMAN. 

The Yorkshire College, Leeds, 
Aprils 1896. 



AUTHOR'S NOTE 

The author takes this opportunity of expressing his 
thanks to Professor Goodman and Mr. Frederick 
Grover, A.M.I.C.E., of the Yorkshire College, for their 
kind assistance; and also to Mr. P. Nicholls, B.Sc, 
Wh.Sc, for the great care with which he has read and 
revised the proof-sheets. 



PREFACE TO THE SECOND EDITION 

The favourable verdict which has been passed on this 
little book by many eminent teachers, and by the 
technical and scientific press (which the author grate- 
fully acknowledges), has given the author more confidence 
than he would otherwise have had in the preparation of 
a second edition. 

The book has been carefully revised, but beyond a 
few corrections, and the addition of several notes, em- 
bodying further and fuller explanations of fundamental 
conceptions, together with a description of the improved 
form of the author's integraph, no substantial alteration 
has been found necessary. 

One of the weightiest criticisms which has been 
passed on the book is that there is a dearth of symbolical 
examples to be worked out by the student, but the 
author respectfully submits that this criticism is due to 
a misapprehension of its aim. That aim is partly to 
form a connecting link between the calculus and other 
branches of mathematics, and partly to enable the 
student to think about the meaning of the calculus from 
the commencement of his study of the subject just as 



x Preface to the Second Edition. 

he can think about the meaning of trigonometry or 
co-ordinate geometry. It is not designed to give him 
facility in symbolical manipulation, for there are already 
a large number of excellent text-books written for this 
purpose. 

The author holds the view that the training of a 
technical student should consist in being taught how to 
think about concrete things rather than about abstrac- 
tions, and though it is possible that this method may 
hinder for a time higher flights into the realm of 
symbols, this may perhaps be not altogether a dis- 
advantage for those who are to be engaged in practical 
life. For these the first essential is a firm grasp of the 
underlying principles of this subject, rather than facility 
in manipulation. The latter may, and does sometimes, 
lead in practice into the most fantastic excesses, while 
the former must invariably restrain any tendency in this 
direction, and is furthermore of very great positive value. 
The author has tried, however, to guard against any 
mathematical inaccuracies of thought or expression 
which would form a stumbling block to those studying 
the calculus simply as an intellectual exercise. 

He has to express his thanks to Mr. J. W. Parry 
for having carefully read the book and pointed out 
several errors and defects, which have in the present 
edition been rectified as far as possible. 



Trowbridge, 

December, 1901. 



CONTENTS 



PAGE 
CHAPTER 

I. Introductory.— Curves and their Equations . . i 

II. Graphical Differentiation and Integration . . 12 

III. Nomenclature and General Principles .... 3 1 

IV. General Principles 4 6 

V. General Principles— (Continued) 63 

VI. Differential Coefficients of Trigonometrical 

Functions 79 

VII. Differential Coefficients of Logarithmic Func- 
tions 96 

VIII. Differentiation of a Function of a Function of 

a Variable with Respect to that Variable . . 105 

IX. Integration I22 

X. Methods of Integration 136 

XI. Miscellaneous Applications of Differentiation . 148 

XII. Miscellaneous Applications of Integration . . 166 

Appendix— Barker's Planimeter 187 

Notes , , , * ,,,,,,,.«.«•• 19 1 



GRAPHICAL CALCULUS 

CHAPTER I. 

INTRODUCTORY, — CURVES AND THEIR EQUATIONS. 
§ I. CO-ORDINATES OF A POINT. 

The exact position of a point in a plane is completely known 
if its perpendicular distances from two intersecting lines in that 
plane are known. Thus, suppose the lines OX, OY (Fig. i) 
represent to some scale two hedges of a field meeting at right 
angles, and we are told that an article is buried in the field at 
a given depth at a point A, whose perpendicular distance from 
the hedge OY is 20 yards, and from OX 30 yards. 

It is clear that the position of the article could be im- 
mediately found by measuring 20 yards from O along OX to 
L, and 30 yards along LA perpendicular to OX. 

Definitions. — The lines OL, LA, would be called the co- 
ordinates of the point A, with reference to the axes OX, OY ; 
OL is the abscissa, and LA the ordinate ; and the point A 
would be described in mathematical language as the point 
whose abscissa is 20, and whose ordinate *o, or shortly as 
" the point (20, 30)." 

§ 2. Equation to a Line. 

Suppose, however, we are told that the distance of the 
point A from OX is half (its distance from OY) + 20 yards. 
From this condition alone we could not find the exact 

B 



2 Graphical Calculus. 

position of the point, for there are many points in the field, 
in addition to the point A, of which the statement would be 
equally true. Thus if we take OM = 50 yards along OX, 
and MB at right angles = (J X 5 o + 20) yards = 45 yards, 
we should find a point B which would also " satisfy the con- 
dition " that its distance from OX was half its distance from 
OY +20 yards. Or we might have taken ON = 80 and 




Fig. i. 

NC = 60 ; or, indeed, any arbitrary (or, as it is called, " in- 
dependent ") distance along OX, and calculated and measured 
the corresponding distance perpendicular to OX. We could 
thus find any number of points " satisfying the given condition. " 
All these points would be found to lie on a certain straight 
line in the field, and no point which is not on the straight 
line would be found to satisfy the condition. 1 And, further, 
any point which is on the line will satisfy it. 

We should then be sure of finding the buried article, if we 
were to dig a trench of the given depth along a line repre- 
sented by AC. 

Now let us attempt to discover the position of the article 
by an algebraical process. 

1 This statement should be tested by plotting the points to scale on a 
plan of the field. 



Introductory. — Curves and their Equations. 3 

Let x be the perpendicular distance of the buried article 
from OY ; andj/ the perpendicular distance from OX. 
Then we have — 

y - - + 20 
2 

This is the only equation we can obtain from the data, 

and we are here met with the same difficulty as before, 

namely, that there are an infinite number of possible solutions 

to the equation, each solution corresponding to one particular 

point on the plan, Now, just as (x = 20, y = 30) may be 

x 
taken to represent the point A, so the equation y = -+ 20 

may be taken to represent the line AC. In other words, the 

line AC is a picture or geometrical representation of the 

x 
equation^ = — h 2 °- To put it in still another way, the line 

AC shows the relation between the value of x and that of 

( — |- 20 ), or y, corresponding to all values of x or y. Thus 

suppose we wish to find from the diagram what is the value 

of y or ( - + 20) when x is 59*2, say, or any other arbitrary 

value, we measure of! to scale 59*2 along OX, and erect a per- 
pendicular to OX from the point so found. The length of this 
perpendicular cut off by the line AC gives the required value. 
The algebraical counterpart of this process is as follows : 

x * 
In the equation y = -+ 20, find the value of y when x is 

59*2. To solve this we have merely to substitute 59*2 for x in 

the equation, and solve the resulting simple equation in y ; we 

thus find the value of y corresponding to x = 59*2. This is 

easier and more accurate than the graphical process. Another 

x 
example of the same thing is, " Find where the line_y = — f- 20 

cuts the axis of y" At the required point it is obvious that 



4 Graphical Calculus. 

x = o. Hence substituting this value of x in the equation, 
we obtain a simple equation in y, viz. y = 20, which gives 
the distance of the point from OX. 
This is expressed by saying that — 

x 

y = - + 20 

2 

is the "equation to the line AC." 

x 
A little reflection will show the student that y = - + 20 

2 

is also the equation to the continuation of the straight line AC 
in both directions anywhere in its length, and not only of that 
part of it which is to the right of OY and above OX. 

Convention of Signs. — In this connection it must he 
noticed that if distances to the right of OY are called positive, 
those measured to the left are to be called negative. Thtf^f 
if a distance =10 yards be measured to the left of OY, its 
distance from OY is said to be — 10. The reason for this may- 
be gathered from consideration of the following case : — 

Suppose a man starts from P to walk to Q, a distance of 4 miles. 
After walking to S, he turns back and walks in the other direction for 

2J miles to R. The total distance, 

h~ 1 H* — ; * a | g ** H irrespective of direction, which he 

R p S 9. has now walked is ij + 2 J = 4 

FlG - 2 - miles ; but, considered with respect 

to his original destination, the effective distance he has walked is — 1 mile, 

i.e. he has I mile to make up before he begins to be any nearer to Q 

than he was when he started. This may be conveniently expressed by 

prefixing the negative sign to distances walked towards the left ; thus, 

iJ + (-2j)=-i, which represents the total distance he has walked 

towards Q. 

Suppose, then, we take a distance along OX = -jo. The 
corresponding value of y will be h X (—10) + 20 = +15. 
The point D (—10, +15) is also on the line AC. 

In the same way distances below OX are reckoned as 
minus quantities. Thus when x = — 50, y = ^ x ( — 50) + 20 
= — 5, the point E ( — 50, —5) being also on the line AC. 

Example. — Find where the given line cuts OX. 



Introductory. — Ctcrves and their Equations. 5 



§ 3. Equation to a Curve of the Second Degree. 
Suppose now that, instead of the condition y = - + 20, we 

had had the condition that the distance from OX = ^of (the 
square of the distance from OY) + 20 yards. 

Dimensions of Quantities.— -It may be noted, in passing, that, strictly 
speaking, it is as absurd to speak of one distance being = the square of 
another distance, as this expression is usually understood (7. e. that a line 
is equal to an area), as it would be to 
say, for instance, that a square foot is 
equal to 6*2 gallons. It is an absurdity 
of the same kind as is often committed 
in mechanics, when speaking of an ac- 
celeration of so many feet per second, 
instead of so many feet-per-second per 
second. Conventionally, however, it is 
to be understood in the following 
sense : We know by geometry that MP 2 (Fig. 3) = AM. MB. Now, 
when MB = 1", there are as many square inches in the square on MP as 
there are linear inches in AM. And so, disregarding dimensions ', we say 
that AM = MP 2 . For instance, if MP were = 3", AM would = 9". 
This idea may also be expressed by saying that we are to regard a line 
as a geometrical method of representing a number, and not necessarily a 
number of inches. Thus a line 3" long represents essentially on the simple 
inch scale the number 3, and may at pleasure stand for 3 seconds, 3 
degrees, 3 feet per second, 3 square inches, or 3 units of any kind what- 
soever. 

Our relation expressed algebraically is — 




y 



40 



+ 20 



Exactly as before, any value may be arbitrarily assigned to 

( ^ \ 

x, and the corresponding value of y or ( — + 20 ] calculated. 

Thus, if x = 10 — 



and so on. 



100 , 
y = — • + 20 = 22-t; 
40 

Or, x = 20 gives y = 30 

x= 30 gives y = 42*5 



6 Graphical Calculus. 

All these and similarly calculated points will be found to 
lie on a certain curve (Fig. 4), instead of, as before, on a 

straight line, and no point 
\ which is not on the curve 

will satisfy the equation, 
and any point which is on 
the curve will satisfy it. 

This equation, since it 
contains the second power 
of one of its "variables," 
o "x is said to be of the second 
degree, and the curve is, as 
before, called "the curve 

y = — + 2 °>" and the diagram of the curve exhibits graphi- 
cally the relation between x and y, or, in other words, between 

q 

x 
x and - — h 20, for all values of x. 
40 

The equation to any curve, then, gives a relation which 
must be " satisfied " by the co-ordinates of any point on the 
curve. We have " plotted " or "traced" these curves by 
arbitrarily assigning a series of values to x (i.e. treating x as 
an independent variable quantity which can assume any value 
we please), and calculating the value which y (the " dependent 
variable ") assumes in consequence of x having that particular 
value we have assigned to it. 

In this book the values of the independent variable are, 
to avoid confusion, always measured in a horizontal direction, 
and those of the dependent variable vertically. 



§ 4. Experimental Curves. 

Curves may also be obtained by other means than trans- 
lating an algebraical equation into geometry, which is practically 
the way in which we have obtained the preceding curves. 



Introductory. — Curves and their Equations. J 

The results of series of experiments in physical or engineer- 
ing science are, whenever possible, plotted on paper. 1 

This method exhibits relations between mutually dependent 
and therefore simultaneously varying quantities far more 
clearly than rows of figures or pages of symbols can possibly do. 
The method may be best explained by taking a simple and 
familiar example. Suppose a kettle of cold water is set on a 
fire, and the temperature of the water observed at intervals 
of, say, one minute by means of a thermometer, a note being 
taken of the time and simultaneous value of the temperature. 
Suppose the initial temperature of the water is 6o° Fahr. 

Our readings are booked as follows : — 



Time (minutes). 


Temperat 


ure (degrees). 


After 


o 




6o 


>> 


i 




95 


>> 


2 


. 


126 


)> 


3 




152 


?> 


4 





174 


?> 


5 





r 93 


>> 


6 




209 


n 


7 




212 



Mark off along OX (Fig. 5) equal distances 1, 2, 3, to repre- 
sent minutes, and along OY (to scale) temperatures beginning 
at any convenient temperature. Then find a series of points 
on the paper whose abscissae represent the observed times, 
and whose ordinates represent the corresponding temperature to 
any assumed scale. Thus the point (1 min. 95 ) will be found 
by the intersection of a vertical through 1 min. and a horizontal 
through 95 . All the points are to be found in the same 
way, and a freehand curve drawn through them. This will be 
a time-temperature curve, exhibiting the result of our experi- 
ment very clearly ; for not only does the height of the curve 

1 The paper most convenient for this purpose is what is called " squared 
paper," which is ruled in small squares of O'l inch side, and can be bought 
at anv stationer's. 



8 



Graphical Calculus. 



at any point show at once the value of the temperature at the 
corresponding time, but the shape of the curve conveys at a 

glance a general idea 
how the rate of rise 
of temperature varies 
throughout the experi- 
ment. It is clear that 
when the temperature is 
rising rapidly (as it will 
do at first), the curve is 
steeper than when it is 
rising less rapidly later 
on in the experiment. 
Thus the amount of rise 
of temperature between 
say o and i min. is greater than the amount of rise between 3 
and 4 min., and therefore the average upward slope of the 
curve between o and 1 min. is greater than the upward slope 
between 3 and 4 min. 

Curves of this kind are quite familiar to us. Thus we 
have in the daily papers curves representing time variations 
in the height of the barometric column or of the thermometer. 
The differential calculus is chiefly concerned w r ith the slope 
of curves, and it is therefore important that we should get 
accurate ideas of how this slope is to be measured. 



Y 

217 


















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60 


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Fig. 5. 



§ 5. Methods of measuring the Slope of a Line. 

Suppose we have a sloping line AB, and we wish to 
determine exactly how much it slopes with respect to a 
horizontal AC. We can do this in several ways, any of which 
we can use when convenient. 

(i.) We can find the number of degrees in the angle BAC 
by a protractor. 

(ii.) We can measure the length of the arc CD by a steel 



Introductory. — Curves and their Equations. 



tape., and also find the length of AC. By these two measure- 
ments we could easily reproduce the angle on paper. 

(iii.) The method we shall 
always use in the differential 
calculus is as follows : From 
any point B in AB drop a 
perpendicular BC to AC. 
Measure CB and AC. Divide 
the length of CB by that of 
AC. We thus obtain the length 
of a line MP where AM = i." 
Thus, suppose CB = 2", AC 

= 3 . Then -7-7. 




Fig. 6. 



MP = 



= 0667. It is clear, if AC is three 
AC 3 

times AM, that therefore CB is three times MP, and therefore — 

CB _ CB 

3 ""AC 

(Read again note on p. 5.) 

This is, of course, true wherever we take B on the line AB. 

If AC, instead of being 3", had been 0*042" suppose, we should 

0*028 2 

0*042 - 3' 



have found CB = 0*028, and therefore MP 



■ = -, as 



before; so that wherever B may be on AB we can always 

CB 

represent the ratio — by the line MP, where AM = 1, even 
AL 

though B is quite close to A. The student should convince 
himself of the reality of this result by trial and accurate measure- 
ment. To obtain the length of MP with some accuracy, it is 
convenient to make AC = 10". Thus in this case CB would 
be 6*67. Hence MP = 0*667. 



§ 6. Trigonometrical Ratios. 
It is convenient to have a name for the ratio 



perpendicular. 
base. 



It is called the "tangent of the angle of slope," and the 
relation is expressed thus in Fig. 6 — 



10 Graphical Calculus. 

Tan BAG = ~ = MP 

AL 

There are other ratios of an angle whieh must be perfectly 

familiar to the student before he can make any considerable 

CB 
progress in this subject. 1 Thus — is called the " sine of the 

AB 

angle BAC," written thus — 

Sin BAC = ^ 
AB 



also — is called the " cosine of BAC," written thus— 
AB 

Cos BAC 

AB 



AC 

is called the " cosine of BA( 

AC 

also " cotangent of BAC," written "cot BAC," = — 

AB 
" secant of BAC," written "sec BAC," = ^ 

, AB 
"cosecant BAC," written "cosec BAC," 



also — 

length of any arc CD 
length of its radius AC 



CB 



= circular measure of the angle BAC ; 



or, in other w r ords, the number of " radians " in that angle. 

It is clear that when the arc CD, measured with a flexible 

steel tape along the circumference, = radius AC, the ratio 

arc CD 

— — — =i. In that case the angle BAC = i radian = 

radius AC 

57-30° about. 

It is clear that, provided CB is always perpendicular to AC, 

all these ratios are quite independent of the position of C on 

the line AC, for as AC increases, BC and AB and the arc CD 

1 It is highly desirable that a student should have a knowledge of 
elementary trigonometry before commencing this subject. 



Introductory. — Curves and their Equations. 1 1 

also increase in the same ratio, if the angle BAC remains 
unaltered. 



EXAMPLES. 

Draw the following curves on both sides of the axes :— 
(i.) y = x. 
(ii.) y = 2x. 
(iii.) y = 2x + 3. 
(iv.) zy - x - 6. 
(v.) y : 



(vi.) y 2 = x + 4. (Notice the double sign for j : y = ± aJx + 4). 
(vii.) y-x 2 - 4. 
(viii.) xy — 4. 
(These curves may be obtained by giving arbitrary values to either x or y.) 

2. What are the meanings of m and c in the line_i/ = mx -{• c? 

(Am. m is the tangent of the angle of slope of the line to OX. c is 
the distance from O, where the line cuts OY.) 

3. Find where the curves 

(i.) 2y = x + 3 
(ii.) y = x 2 - 2 
(iii.) y 2 = x - 3 

cut the axes of X and Y. 

(Find the value of x andj successively when the other variable = o.) 

4. Find the equation to a line cutting OY at a distance = 3 below O 
and inclined to OY at 6o°. (See Example 2.) 

5. Why is no part of the curve _y = x 2 below OX ? 

(Am. Because, whether the value of x be positive or negative, the square 
of it must essentially be positive, i.e. the ordinate must be above OX. Sup- 
pose any point of this curve were 2 inches below OX ; we should have — 

— 2 = x 2 

or x = aJ — 2 

And since the square root of a negative quantity is essentially imaginary, 
being neither -f- nor — , it is evident that we can have no real point of the 
curve below OX.) 

6. Devise geometrical constructions for determining the value of the 

sine, cosine, circular measure of any angle : also for -. — , — : — , etc 

** sine cosine 

(See iii. p. 9.) 



CHAPTER II. 

graphical differentiation and integration. 

§ 7. Illustration of Graphical Differentiation. 
Every straight line, however long or short, must have a 
definite inclination to every other line in its plane. In this 




Fig. 7. 

work we are chiefly concerned with that function 1 of the 

1 Any variable quantity p whose value depends on the value of another 
variable quantity q, is said to be a " function " of that quantity. Thus the 
sine of an angle is a function of that angle, because the value of the sine 
depends on that of the angle. 



Graphical Differentiation and Integration. 13 



LEVEL 



inclination of lines to the horizontal which we have called 
the tangent of the angle of slope (§ 6). 

If we have a figure made up of straight lines, it is quite 
easy to determine the slope of each rectilinear part of it to a 
horizontal line. Suppose, for instance, we have given an 
elevation (drawn to scale) of a certain section of railway, such 
as Fig. 7, in which vertical heights are much exaggerated for 
the sake of clearness. 

On most railways what are called the " gradients " are 
indicated on boards, such as that illustrated in Fig. 8, placed 
by the side of the line. The 
meaning of this is that w T hile 
the line is level on the right of 
the board, for every 100 feet 
measured horizontally, Howards 
the left the line falls 1 foot , 
expressed in other w T ords, the 
tangent of the angle of slope 
is Y^o- The direction in which g 

the line slopes is shown by the 

obvious slope of the board. If the line is level, or of no 
slope, it might in the same way be indicated o in too, but the 
word " level " is used instead. 

Now, we may very conveniently draw underneath the actual 
elevation of the railway in Fig. 7, another curve showing the 
slope at each point ; or, choosing the scale of 1 inch = 1 in 100, 
the rate at w 7 hich the line is rising just at that point for every 
100 feet horizontal. It is of the highest importance that the 
student should thoroughly grasp the exact meaning of this 
lower curve, for in it is contained the very kernel of the whole 
subject. Suppose we take any point A on the upper curve 
(which w r e may call a curve in spite of the fact that it consists 
wholly of straight lines). 

Draw a vertical from A to cut the lower curve in A\ Then 
the height of the lower curve a x A y shows the amount (1 foot) 

1 See note (1), p. 191. 



14 Graphical Calculus. 

which the line would rise if it continued with the same 

slope as it has at A for ioo feet horizontally. 1 The fact 

that the slope may change just after the point A is passed 

does not in the least affect the height just at the point 

A\ for this is only dependent on the tangent of the angle 

of slope just at the point A. Neither is the height of the 

lower curve at A K any guarantee that the line will actually rise 

i foot in the next ioo feet horizontal, any more than the fact 

that a train may be travelling at the rate of 60 miles per hour 

is any guarantee that in the next hour it will actually travel 

60 miles, or any other distance. It possibly may entirely 

change its velocity in the next half-second, as in the case of a 

collision, but this does not alter the fact that at the point in 

question it was travelling at 60 miles per hour. In exactly the 

same way, at the point B, the line is rising at the rate of 1 foot 

per 100 feet, and this is not affected by the fact that from a 

couple of feet to the right the line runs perfectly level. The 

important point to observe is that the height of the lower curve 

represents a rate of rise, and not necessarily an actual rise. It 

. _ . . corresponding small vertical rise. 

represents, in fact, the ratio — - — : — 

small horizontal distances. 

The fact that both the numerator and denominator of this 

fraction may be indefinitely small does not affect the value of 

the ratio, as explained in § 5. 

Now, just to the right of point B the upper curve suddenly 
changes its slope from 1 in 100 to o in 100; and, conse- 
quently, the lower curve drops suddenly from 1 to o. The 
line has no slope from near B to near C, and therefore the 
lower curve has no height between the same two points. 
Near C the rise changes into a fall, i.e. the rate of rise 
becomes negative, and so the height of the lower curve becomes 
negative, i.e. is below the axis of X (see § 2). At D the negative 
slope suddenly changes to a positive one, and so the lower 
curve suddenly jumps up above the axis. 

If the student has thoroughly mastered the preceding 

1 See note (2), p. 191. 



Graphical Differentiation and Integration. 15 

explanation, he will have little difficulty with the rest of the 
subjects treated of in this book, and we have, therefore, given 
it a very full explanation — fuller, perhaps, than many students 
will find necessary. The principle here explained is exactly 
that running through the whole of the subject, and the student 
cannot be too familiar with it. The lower curve is called the 
" derived curve " of the upper one, and the height of the lower 
curve at any point gives the value of the " differential co- 
efficient " of the upper one at the corresponding point. The 
student will understand these expressions better a little later 
on. We shall then also see that the railway companies, by 
means of boards, such as Fig. 8, really " differentiate " the 
curve of the railway for the information of the engine-drivers. 

§ 8. Example of Graphical Integration. 

Hereafter, in all cases, points on the derived curves are in- 
dicated by dashes, thus : P y o?i the first derived and P" on the 
second derived correspond to P on the primary, etc. 

Let us now look at the converse process. Suppose we are 
given this derived curve or curve of slopes, and are required 
to deduce from it the actual elevation of the railway. The 
student will find this easy if he has mastered the principle on 
which the derived curve was obtained. We see, in the first 
place, that along the line PB there must be an even uphill 
slope of 1 in 100; for the fact that P r B r is parallel to 0\X* 
shows that the slope is constant. Hence PB must be 
perfectly straight as far as B. A question that meets us at 
the outset is a very suggestive one. Where are we to start to 
draw the curve ? The bearing of this question on the subject 
of integration will be fully explained later on. At present it 
is sufficient to notice that there is nothing whatever in our 
derived curve to tell us where the point P is to be taken in 
the vertical 0^0 ; that is to say, our curve of slopes does not 
give us the height of the point P, or any other point, above 



1 6 Graphical Calculus. 

datum level. Taking, then, any arbitrary point P as starting- 
point, we see that the line slopes upwards i in ioo as far 
as a point corresponding to B\ after which it is level as far 
as a point vertically above C\ afterwards sloping downward 
£ in ico as far as D, then upwards ^ in ioo. So that we 
can draw the actual shape of the upper curve, having given its 
curve of slopes. 

Meaning of the Arbitrary " Constant" in Integration. — 
If we had taken any other point, P lf as starting-point in the 
same vertical line, we should have obtained the dotted curve 
which is precisely similar to the one we have obtained, but shifted 
higher or lower, according as F 1 is higher or lower than P, and 
the distance between the two curves, measured perpendicularly 
to OX, is " constant " all along the curve. It is clear that 
while we cannot find the absolute height of any point on 
the curve so obtained, yet we can ascertain definitely the 
difference of height of two points on it, for this difference is 
the same wherever the curve, as a whole, may be shifted to. 

§ 9. Differentiation of Continuously Varying 
Outline. 

It is not necessary, for our process, that the real elevation 
of the curve should consist of straight lines. We might take 
any continuously curving outline, such as the hill illustrated 
in Fig. 9, and determine the curve of slopes for it. Now, in 
a rounded outline like this, the slope does not change suddenly, 
as it did in our assumed case of the railway, but it gradually 
changes from point to point. We say familiarly that the hill 
is steeper at one part than at another, or that as we go higher 
up the hill it gets steeper and steeper, or the steepness 
increases every step we take, and so on. Now, we have 
already explained how we are going to measure this steepness, 
viz. by the tangent of the angle of slope (§ 5). But we come 
again to a question which must be carefully considered in all 
practical applications, viz. what scale are we to use ? Observe 



Graphical Differentiation and Integration. \J 

that the horizontal scale is the same for both curves, whereas 
a vertical height or ordinate on the upper curve has an entirely 
different meaning from an ordinate on the lower ; the former 
represents an actual height, the latter a ratio. For con- 
venience and clearness, we shall at present adopt as a unit 
whatever is represented by i" on the upper figure/ which is 




Fig. 9. 



here 100'. The vertical scale of the derived curve is here 
full size where PM is unit length, i.e. the height in the given 
units represents the tangent of slope at the corresponding 
point of the upper curve. 

To find the derived curve, draw a number of ordinates 

1 In the original drawing, of which Fig. 9 is a reduced copy, PM was 
made = 1". This remark applies to all the figures. 

C 



1 8 Graphical Calculus. 

to the upper curve and produce them downwards. We are 
about to determine the actual slope of the hill at each of the 
points where these ordinates cut the outline of the hill. For 
the sake of avoiding confusion in the figure, we shall do this 
for two points only, viz. P and Q ; all the others are to be 
treated in exactly the same way. At the point P draw a line 
PT, touching the curve. The slope of the hill at the point 
P is evidently exactly the same as that of this line, for a little 
to the left of P the hill slopes more, and a little to the right 
less ; so that at the point P the slope is the same. Through 
P draw PM horizontal = i" (representing 100'), and draw 
MT vertical. This line measures, say, 1*2" =120 feet. 
Then, just at the point P the hill is sloping 120 feet in 100. 
Take any convenient base-line, O^X\ on which to draw the 
derived curve, and make p y P y = MT ; then ^P x represents the 
slope at the point P. At Q the hill is sloping downwards. 
Draw the tangent as before, and make QN = 1", and draw 
NS vertical, and make ^Q* on the lower curve = NS, and 
measured in the same direction, i.e. downwards. Repeat this 
process for all the points where the ordinates cut the outline. 
It is convenient, before commencing, to ink in the original 
curve, so that any line may be removed from the figure after it 
is done with. It is usually more accurate to draw the tangent 
first, and the ordinate afterwards ; the point of contact can 
then be more accurately found. The drawing should be made 
with a hard pencil, sharpened to a fine point. Unless the 
original curve is very accurately drawn, it is impossible, even 
with great care, to obtain very accurate results, owing to the 
difficulty of drawing in the tangents correctly. When the 
points are all obtained, draw an even curve through them all. 
This is the derived curve, or curve of slopes of the upper 
curve. The height of it represents, at any point, the corre- 
sponding value of the differential coefficient of the function 
which represents the height of the upper curve. 



Graphical Differentiation and Integration. 19 

§ 10. Remarks on Derived Curves. 

There are several important facts to be observed about 
two curves standing in this relation to one another. 

1. At the highest point of the hill the slope is, of course, 
nothing, the tangent being horizontal \ for suppose the slope 
were slightly downwards, then a point on the left of the 
highest point would be slightly higher than the highest point, 
which is absurd, (This assumes that the curve is continuous, 
i.e. that there are no angles or sharp points in it.) The height 
of the derived curve is therefore nothing. 

2. This is also the case at the very bottom 
for a similar reason. 

3. Conversely, at the point corresponding 
the derived curve cuts the axis of X\ the 
primary curve is either a maximum or minimum. 

4. In the case of a maximum, the derived curve slopes 
downwards from left to right, i.e. has a negative slope. In the 
case of a minimum, the derived curve slopes upwards from 
left to right, i.e. has a positive slope. 

By a "maximum" or "minimum" we do not mean absolutely the 
highest or lowest point on the curve, but merely a point to the left of 
which the slope is in a different direction from what it is on the right. 




Fig. 10. 



Thus in Fig. 10 the points AAA are all maxima, and BBB are 
all minima. For each of these points the derived curve crosses its own 
base-line. 



20 Graphical Calculus. 

5. At the point where the curve of the hill changes from 
-convex to concave, or vice versa, with respect to the axis OX 
its slope is greatest, and consequently the derived curve highest 
or lowest, as the case may be ; i.e. at these points the derived 
curve has no slope (see observation 1). Such a point is called 
a " point of inflection " on the original curve. 

§ 11. Meaning of a Tangent to a Curve. 

We have spoken of " drawing a line touching a curve," but, 
as much of what follows depends on the relation between 
< curve and its tangent, it is important to get clear ideas as 
\ f-he precise meaning of the expression, "a line touching 
rve." Euclid's definition applied to a circle is that the 
line meets the curve, but, being produced, does not cut it. 
This definition, although applicable to a circle, would not be 
applicable, for instance, to such a curve as that illustrated in 
Fig. 10, where the tangent at P cuts the curve again at S. 
The following is a more modern conception of a tangent to 
a curve. Consider a fixed point P and a movable point Q x 
on a continuous curve of any shape. Join PQ 1? and produce . 
it in both directions. Conceive that Q, with the line PQ 
always passing through it, moves gradually towards P, occu- 
pying successively such positions as PQ l5 PQ 2 . " In the limit " 
when Q is just on the point of coinciding with P (being, in 
fact, " infinitely near" to P), the line PQ is a tangent to the 
curve at P. In this position the infinitely small portion of 
the curve PQ may be regarded as coinciding with the portion 
PQ of the touching line. This shows that a curve may be 
regarded as composed of an infinite number of indefinitely 
short straight lines joined end to end. Bearing this explana- 
tion in mind when we have to do with an extremely short 
bit of a curve, we shall treat it as a straight line, as the 
various explanations are thereby rendered much simpler and 
clearer. 



Graphical Differentiation and Integration. 21 

§ 12. 1 

We have also a remark to make with respect to the modern mathe- 
matical conception of the meaning of such expressions as "zero," 
"infinitely small," "infinitely great," "absolute equality," and so 
on. It may be stated at once that the human mind is incapable of 
conceiving any reality corresponding to any of these expressions. The 
modern conceptions of them may be briefly summed up thus : zero or o 
means " something smaller than anything" Thus, take a quantity whose 
weight is the thousand-millionth part of the weight of a hydrogen atom ; 
zero weight means some ivcight smaller than this— smaller, indeed, than 
anything that can be named. The same idea is contained in the words 
" infinitely small." In the same way, "infinitely great " means "some- 
thing greater than anything." " Absolute equality " between two quantities 
exists when the difference between them is what we have defined above 



§ 13- 

Let us now cease to regard the upper curve of Fig. 9 as the 
elevation of a hillside, and look at it simply as a curve traced 
on paper. The lower curve may still be called the curve of 
slopes of the upper one. The length of the ordinate of the 
lower curve, corresponding to a point P on the upper, now 
simply shows the instantaneous rate at which the upper ordinate 
is increasing per inch increase of the abscissa, i.e. the rate of 
increase during the instant in which the tracing point is 
passing through P ; of course, after the point has passed 
through P the rate is no longer the same. 

Comparing together § 5, § 9, and § 12, we see that if, in 
Fig. n (which represents a curve PB and its " first derived"), 
P and Q are two points on the primary curve close together, 

KQ 

then z- — : = p y F y (This is only an absolute equality when Q 
PK 

is infinitely near to P.) Therefore, since pq is small, KQ 

= /FxPK=/Fx/V\ nearly. That is to say, the 

number of linear inches in the short line KQ is very nearly 

equal to the number of square inches in the thin rectangle 

PKV/\ In the same way, let pq = qr = rs 9 etc. ; 

1 See note (3^, p. 191. 



22 



Graphical Calculus. 



Then LR = rectangle Q'LVy 

MS = RW1VV 

and so on. Suppose this process continued as far as B, and the 
results added together; then clearly to the same approximation — 

KQ -f LR + . . . + HB in inches = P*Ky/ + Q v LVy + . . . 

+ ITHW in sq. inches (a) 

Now,- however long or short pq,qr, etc., are, the left-hand 

side of this equation always = CB. Imagine, then, that instead 

of pq, qi\ etc., being finite, as 
in the figure, we had been able 
to draw the ordinates /P, ^Q, 
etc., so close together that pq, 
qr, etc., are infinitely small, so 
small that pb contains an in- 
finitely great number of small 
parts, each - pq. This will 
not in the least alter the total 
value of the left-hand side of 
equation (a), for this must of 
necessity be = CB. But when 
pq, qr, etc., are infinitely small, 
the right-hand side is equal to 
the area under the curve P\B\ 
bounded on each side by the 
ordinates p^ and PB y \ for it 
is clear that the sum of these 
rectangles only differs from this 
area by the sum of the little triangles P r *CQ\ Q^R\ etc., and 
these all added together are evidently considerably less than 
the rectangle K^D, which is infinitely small compared to the 
area P\B^ A , when p y <p becomes infinitely small (see § 32 on 
" Orders of Infinitesimals "). 

So we see that the dwindling of pq to an infinitely small 
quantity produces two effects simultaneously — 




Graphical Differentiation and Integration. 23 
(1). Makes the equations such as 

PK F 
absolute equalities instead of only approximations. 

(2) Makes the equation — 

Sum of rectangles P^Ky^ -f- etc. = area of curve 

an absolute equality instead of an approximation. 

On the other hand, it does not interfere with the absolute 
equality KQ + LR + . . . = CB. It therefore makes equation 
(a) equivalent to the assertion — 

CB in inches = area P^B^y in sq. inches . (/3) 

(see note to § 3 and § 12). 

This result must not be regarded as an approximation. It 
is an absolute and complete equality. 

Our proof of the equality (j8) would have been only an approximate 
one if we had imagined the strips as of finite thickness ; for in that case 
the two equations on which (j3) depends would de only approximately true. 
They are only absolutely true (§ 12) "in the limit " when pq t etc., are 
infinitely small. 1 

Here, then, we have a most convenient way of measuring 
any irregular area with a curvilinear outline. Suppose, for 
instance, it is required to find the area of the lower figure 
between any two ordinates. we must regard it as the curve of 
slopes of some other curve, which we must proceed to draw. 
We shall presently explain the practical method of drawing 
this curve. Assuming that it has been drawn, and that the 
upper curve in Fig. n has been so obtained, then, in order 
to find the area of the lower curve between any two ordinates 
taken at random, we have only to find the difference between 
the two corresponding ordinates of the upper curve, and we 
have at once the required area in square inches. 

1 Read carefully note (4), p. 192. 



2 4 



Graphical Calculus. 
§ 14. Practical Integration. 



The method of finding the upper curve, having given the 
lower one, is the same in principle as that by which, in § 8, 
we deduced the curve of the railway from its curve of gradients. 




Take a curve, such as the lower one in Fig. 12, of which 
we are desired to find the area between any two ordinates. 
Then we know that the length of any ordinate of this curve 
in inches must = tangent of slope of required curve at the point 
where continuation of that ordinate cuts it. Draw a number 
of continued ordinates, say y apart, and figure them as shown. 



Graphical Differentiation and Integration, 25 



Set off on the axis of X, OS - 1", and set up SK vertical. 
Cut off on SK, Si, S2, etc., etc., equal to the ordinates o, 
!> 2 , 3? 4> respectively. Join Oo, Oi, etc., as shown. Now 
consider any one of these ordinates, say 3. If we had had 
the upper curve given, and had been required to deduce the 
height of the lower on the ordinate 3 by the process explained 
in § 9, it is evident that the triangle w 7 e should have drawn 
would have been exactly equal and similarly situated to the 
triangle OS3. Hence O3 must be parallel to the tangent to 
the upper curve at the point where ordinate 3 cuts it. 
Arguing in a similar way about the other points, it is clear 
that what we have to do is to draw a curve of which the 
slopes at the points where the lines o, 1, 2, etc., cut it, are the 
same as the slopes of the lines Oo, Oi, 02, etc. On pro- 
ceeding to draw the curve, we take an arbitrary point T from 
which to start (re-read carefully § 8 on the " Constant "). 

The next question is as to how we are to obtain the form 
of the curve between two ordinates. This we shall decide as 
follows. Any two neighbouring ordinates will be assumed 
so close together that the part of the 
curve between them is indistinguishable 
from a circular arc which has the same 
slopes at its ends as the actual curve 
would have. It will not be necessary 
for our purpose to find either the p 
actual position of the centre or the 
length of the radius. The method of 
obviating this necessity is seen from 
Fig. 13, which is an enlarged view 
of two neighbouring ordinates. P is 
the assumed starting-point. Draw 
through P, PR, and PS parallel to the 
directions of the curve on the ordinates P and Q. These 
directions are obtained from Oi, O2, O3, etc., in Fig. 12. 
Bisect RPS by PQ. It is evident, from the geometry of the 




Fig. 13. 



26 



Graphical Calculus. 



figure, that the circular arc when drawn will pass through Q. 
Q may next be taken as a fresh starting-point and the process 
repeated, and so on for the next ordinate. Thus a series of 
points are found through which the curve may be drawn. 
The difference between any two ordinates of the upper curve 
will, as already proved, give the area of the lower curve 
between the same two ordinates. 



211 




Fig. 14. 

It is to be noticed that when the lower curve dips below the base, the 
corresponding portion of area is to be reckoned negative. 

This is the process of " integration/' and corresponds exactly 
to the algebraical process known by that name in mathematics. 

It is necessary, for the above process, to use a hard sharp- 
pointed pencil, otherwise great inaccuracies may creep in. 
If carefully performed, the process is at least as accurate as 



Graphical Differentiation and Integration. 27 



any of the ordinary processes for finding of areas. The author 
of this work has devised a mechanical integrator, described in 
the Appendix, whereby the integral curve may be automatically 
drawn. 

§ 15. Differential Coefficient considered as a Rate 
of Increase. 
In cases where the curve which we are differentiating is one 
representing the results of a series of experiments, the derived 
curve is often of great importance. As an illustration of 
this, we will take the case of the experiment described in 
§ 4. Suppose we had found, in that experiment, that the 
temperature had risen uniformly up to 212 — that is to say, 
that, during the first half-minute, if the temperature had 
risen 9^° (suppose), then during 
the fifth, or any other half-minute, 
it would also have risen 9^°. 
Suppose the total time occupied 
= 8 min. Now, if we divide the 
total increase of temperature, viz. 
152 (represented by O212, Fig. 
14), by the time occupied, viz. 8 
min., represented by ON, we shall 
obtain the amount of rise of tem- 
perature in 1 min. We have seen 
from § S that we shall also obtain 
the line MP. It is also otherwise 
obvious that, since OM = 1 min., 
MP = amount of rise of tempera- 
ture in 1 min. Thus the tangent 
of the inclination of OP to OX 
represents the rate at which the temperature is rising. Our 
derived curve in this case would be a line parallel to O y X y 
at a height = MP. This would indicate that the rate of rise 
of temperature was constant all along the curve. 



V 












?yT 





















P 
? 


X 


0' 




P s 




X 



Fig. 15. 



28 



Graphical Calculus. 



Although the actual curve was not a straight line, it may 
easily be seen that the tangent of inclination at a point P 
(Fig. 15) still represents the rate of increase at the point P, 
for the small " element " of curve at the point P is also part of 
the tangent to the curve (§ 11), and the rate of increase is 
therefore the same as the rate of increase along the tangent, 
i.e. = height of derived curve at the point. 

§ 16. Other Applications. 

Innumerable other applications of the same principle may 
be found. In almost every case of a curve derived from 
experiment, a distinct and tangible 
meaning may be ascribed to the 
height of the derived curve. One 
of the most important applications 
is the case where a curve is drawn 







Q 


y 




P 






f 











seconds 


r 


V 


1 




P 


q' 


f 










seconds t 


D> 


9' 



FiG. 16. 




to represent the motion of a moving body. Take the case 
of a man walking at a uniform rate along a road. Suppose 



Graphical Differentiation and Integration. 29 

we plot vertically his distance from the starting-point, and 
horizontally the corresponding time.' Thus after 1 sec. (repre- 
sented by 0/) (Fig. 16) he has walked a distance repre- 
sented bypF ; ;■— = ; — = p y V y - his rate of walking. The 
Op Oq 

derived curve is here a horizontal line, as in the last section. 

Successive Differentiation. — But suppose he walked a 
greater distance in the second second than in the first, and 
a greater still in the third, and so on, the height of the derived 
curve would still represent his instantaneous velocity at the 
corresponding point; for just at the point P, for instance, he 
is increasing his distance from any fixed starting-point at the 
rate of MS = p s V s yards per second, so the middle curve is 
a curve of velocities. Now, if we differentiate the derived curve, 
we shall obtain a curve showing the time-rate at which his 
velocity is varying at each point, for at P x his velocity is 
increasing by IVTS* yards-per-second every second =/"P M . 
Hence p"P yy represents the numerical value of his acceleration 
at the point P. This curve is called the second derived 
curve of the time-distance curve. If we differentiate again, 
we shall obtain a curve showing the rate at which his accele- 
ration is changing. The dimensions of these latter units 
would be yards-per-second-per-second per second. This 
curve is called the third derived of the ti me- distance curve, 
and so on. It is easy to see that if the velocity had increased 
uniformly (or the time velocity curve had been an inclined 
straight line), the acceleration curve would have been a 
horizontal line, or the acceleration would have been constant 
or uniform, as in the case of a falling body. Thus we see 
that velocity = time rate of change of distance, acceleration = 
time rate of change of velocity, etc. 

The student should think very carefully over this argu- 
ment, because, in addition to its intrinsic importance, it forms, 
perhaps, the most perfect illustration of the application of 
the calculus to science that could be found. Curves may 



30 Graphical Calculus. 

sometimes be differentiated by special constructions not in- 
volving the drawing of tangents. Several cases of this will 
be given later on. 

Examples. 

1. The space passed over by a body falling from rest is given by — 

y - 16/ 2 
Draw this curve, selecting suitable scale, and differentiate it twice 
graphically. Compare the curves obtained with the curves (i.) y x = 32/, 
(ii.) y v = 32. What are the meanings of y, y\ y KX ? 

2. Draw also the curve — 

y - lot + \6t 2 

giving the space described by a body thrown downwards with velocity 

ft. 
10 — ; l differentiate it twice, and compare with curves derived from (1). 
sec* 

Show that the first derived differs by a constant from the corresponding 

curve of (1), and the second derived is the same as the second derived of 

(1). What would have been the difference introduced if the body had 

been thrown upwards ? 

3. Draw the curve xy — 12. Integrate it, and find, by means of the 
integrated curve, the area of the given curve between the ordinates — 

x — 3 and x — 12. (Ans. — 16'6$ sq. inches.) 

4. Draw the curves (in the first quadrant only) — 

1 _ 

y~Yo ^ 2 andj= J x 

and find their areas between the ordinates x — 3 and x — 8. Ans. 16*17 
and 1 1 '6. 

5. Find between the same ordinates (by reducing the scale of the 
integrated curve) the area of — 

y z — x 5 i.e. y = xK A?is. 89 sq. inches. 

1 This is a very convenient and suggestive notation for the unit of 
velocity, i.e. foot-per-second. It is clear that a velocity of say 6 feet per 
second is the same velocity as 12 feet per two seconds, or 18 feet per three 

seconds. This notation brings this out very clearly for - obviously 

= = It will be found on examination that any quantity 

2 sec. 3 sec. 

preceded by the word " per " is invariably a denominator. A logical 

extension of the same notation is used to denote unit of acceleration. An 

increase of velocity of s feet-per-second every second is denoted — 

ft. 

sec. ft. 

sec. ~~ sec. 2 
All physical units are treated in the same way. 



CHAPTER III. 

nomenclature and general principles. 

§ 17. to show that the height of the derived 

Curve may reasonably be denoted by — . 

ax 

We have shown, in the last chapter, the geometrical meaning 
of the processes of differentiation and integration. We now 
proceed to explain the system of symbols that accompanies 
it. Suppose we are required to find the height of the derived 
curve of the curve y = x 2 , corresponding to the point (2, 4). 
Now, it would obviously be very inconvenient to be com- 
pelled to draw the curve y = x 2 to scale, and differentiate it 
graphically in order to find the height of the derived curve 
at one point. An algebraical method of calculating it would 
be much more convenient, and this is what we are about to 
explaia 1 The method is as follows : — 

1. Calculate the height of the primary at a point Q (Fig. 18) 2 
whose abscissa is slightly greater than that of P, the given point. 

2. Find the difference MQ between the ordinates, and 
divide it by PM, the difference between the abscissae. 

3. Investigate what the result would be if PM were to 
gradually diminish until it became infinitely small. Now, if 

PM is a considerable size, the value of - — viz. the tangent 

PM' & 

of the angle of slope of PQ, will differ considerably from the 
1 See note (5}, p. 194. 2 Fig. 18 is not drawn to scale. 



32 



Graphical Calculus. 



tangent of angle of slope of the line touching the curve at P. 
When PM is infinitely small, there will, be no such difference. 
Our algebraical method of limiting values is equivalent to 
taking Q infinitely near to P. The algebraical result of the 
process must not, therefore, be regarded in the light of an 
approximation. It is an absolute exact truth (see § 12, p. 20). 




Fig. 18. 

The difficulty in understanding this (if there be one) is 
due to an imperfect appreciation of the meaning of the 
expression, "when Q is infinitely near to P." Now, PM is 
called the " increment of x." It is written A*, and implies 
the amount by which x is supposed to increase from an 



Nomenclature and General Principles. 33 

assumed, particular value x. Similarly, MQ is called the 
corresponding " increment of y" and is written Ay ; 

Ay 
Tan QPM therefore =-~ 

^ A* 

Now, the point P being (2, 4), let the abscissae of Q be 
2^ i.e. A* = h Its ordinate then = (2J) 2 = 6\. 

Hence Ay = {( 2 i) 2 - ( 2 ) 2 } = 2\ ■ 

Thus when x increases by i from the value 2, y increases 

by 2\. 

Ay 
Therefore — = 45 when A* = \ 

A# 

Similarly, if we take A^ = \— 

Ay = i T V 

Av 
and therefore — = 4! when Ax = i 

A* 

Similarly, if Ax - o'oooooi 

A_y 
then — = 4*000001 

Ax 

Again, if we take Ax = — \ — 

Then Ay = {(if) 2 - (2> 2 } = -|f 

A*~ -i 3 * 

If we plot a curve (Fig. 19) showing the relation between 

Av 
Ax (abscissa) and — (ordinate) for the point (2,4), we shall 

find it is in this case a straight line. Where this line cuts 

OY, the value of Ax is obviously = o ; i.e. the increment of x 

(PM in Fig. 18) is "indefinitely small" or " vanishes," and in 

Ay 
this case clearly — = 4 exactly. 

Ax 

When this is the case, i.e. when Ax, and therefore Ay, 

D 



34 



Graphical Calculus. 



diminish indefinitely, we write dx and dy instead of Ax and 

by, and the ratio (as indicated in Fig. 19) — . Bat — is not 

dx dx 

i-i A y • 

like — in being an actual 

fraction with numerator and 
denominator— that is to say, 
the dy and dx are not separate 
quantities which have actual 

numerical values ; — must be 
dx 

taken as a single symbol repre- 
senting a definite finite quan- 
tity, although dy and dx are 
each infinitely small (see 
bottom of p. 9). 

Of course, it is only for the 
point (2, 4) that the value of 




/ i 



1 z 

Values of A a: dy 
Fig. 19. 



dx 



4. If we had chosen the 
point (3, 9) instead, we should, in exactly the same way, have 
6. Indeed, if we had taken any point (a, a 2 ) on 



found -- 
dx 



dy 



the curve, the value of — would have been 2a. 
dx 



The student 



should work out a few cases of this in the same manner as 

shown above. If he does it thoughtfully, he will probably 

be able to see the reason of it. 

The algebraical process corresponding to that explained 

above is as follows — 

Let y = x 2 . . . . . . . (i.) 

Let x increase by Ax, and in consequence y by Ay. 

[In the process explained above, we took particular values J, J, — |, etc., 
for Ax, and calculated the corresponding numerical value for Ay. Here 
we calculate the general algebraical value for Ay in terms of Ax as 
follows.] 






Nomenclature and General Principles. 35 

Then, since the point (x + Ax, y + Ay) is by supposition 
on the curves = x\ we have (see p. 6) — 

y + Ay - (x 4- A.*) 2 . . (ii.) 

Subtracting (i.) from (ii.) in o:der to find the difference 
between the ordinates, we have — 

Ay = (x + Ax) 2 — x 2 — 2XA3: + (Ax) 2 

Dividing by Ax the difference between the abscissae 
(PM in Fig. 18)— 

Ay 

- — = 2x + Ax 
Ax 

which, when Ax is indefinitely diminished, becomes — 

dy 

~ — 2X 

ax 

because Ax (or, as it would then be written, dx) becomes 

an infinitely small quantity. Therefore, as in § 12, the dif- 

dy 
ference between — and 2x being infinitely small, we say that 
dx 

dy 

— = 2x absolutely. 

(The student should compare this process with that ex- 
plained above at every step. Only thus can he fully realize 
its meaning.) 

Therefore the ordinate of the first derived curve, of the 
curve y = x 2 , is always twice the abscissa, i.e. the derived 
curve is a straight line whose equation is y y = 2x. 

Exercise. — In exactly the same way, the student can 
calculate the height of the derived curve for y = x 3 . He will 
find it to be y y = $x 2 ; for y = x A , it is y y = ^x\ 

It will now be evident that the process of finding the algebraical value 

of — is that of obtaining the equation to the curve which shows the 
dx 

relation between -^ (ordinate) and Ax (abscissa), and finding, as at p. 4, 



36 



Graphical Calctihts. 



where this curve cuts the vertical axis. This, it is clear, gives the value 
Ay 



Ax 



when Ax = o. 



We can here prove part of the general proposition that 

This is true, as a matter of fact, 



dy 
when y — x n , — = nx [n ' 1) , 
ax 



whatever n may be, positive or negative, integral or fractional. 
The reader is, however, not yet in a position to understand 
the complete proof, so we shall confine ourselves here to 
the limited case of positive integral indices, which will be 

necessary for purposes of 
illustration. The rest of 
the proof will be given 
as the reader is ready 
for it (§§ 24, 34). 

All complete curves of the 
form y = x n where n is posi- 
tive and > 1, are in general 
x shape similar to two of the 
four branches of the curves 
shown in Fig. 20. 

Where n is an even posi- 
tive integer, the curve y = x n 
resembles the curve PjOP 
and when n is odd, P 2 OP. The 
reason of this is that, whether x 
is positive or negative, y or 
x n is always positive when n 
is positive and even (thus 
(-2) 4 = + 16), so that the 
ordinate y always lies above 
XOX. When n is odd,j/ or x n is always negative when x is negative, so that 
in such curves as y = x 5 the curve on the left of OY always lies below XOX. 

For instance, in the curves = x z , itx = - 2, y = ( — 2) 3 = — 8. The branch 

3 
OP 3 is included in such cases as y 2 = x 3 , i.e. y = + x' 2 . In cases where « 

is < I, but > o, the curves resemble Fig. 20 turned through a right angle, 

i.e. looked at with OY horizontal. When n is < o, the curves resemble 

hyperbolas, of which OY, OX are asymptotes. 

It is a most interesting exercise to trace the variations of the curve 



\ Y 

\p. 


- -/p 


IV 


7\ 


P iL _ 


\ p 


v l h- 


\ 



Fig. 20. 



Nomenclature and General Principles. 37 

y — x\ as /; vaiies between +00 and -00. It affords, among other 
things, a beautiful illustration of the meaning of the statement that 
x° = 1. 

"Ltty = x n . . ... . . (i.) 

If x increase by Ax, and y in consequence by Ay, we have, 

as before — 

y + Ay = (x + Ax) n . . (ii.) 

Subtracting (i.) from (ii.), we have— 

Ay = ( X + kx) n - X n 

Ay __ (x + Ax) n — x n 

Ax Ax 

Expanding by the binomial theorem — 

n(n — 1) 

x n + nx n ~ 1 Ax + — L x n - 2 (Ax) 2 + . . . - x n 

A^ =i __ i-2 v ; _ 

Ax Ax 

= nx n ~ x + Ax X some other quantity 

which equation, when Ax vanishes, becomes f- = nx^^; or, as 

ax 

it may be written — 

r (n-l) 



d(x n ) 

- = nx x ' 



dx 



§ 18. Meaning of dy and dx when used alone. 

In some methods of treating and writing the calculus the 
expressions dy and dx are used apparently alone. This seems 
to cause great difficulty to students, because of a sort of 
indefiniteness in the actual values to be assigned to the 
quantities denoted by dy and dx. It will be found on ex- 
amination, however, that in such cases there is always an 

dy 
implicit reference to the ratio -- . It is merely a somewhat 

more convenient way of referring to the ratio, and is introduced 
for the purpose of saving space and for convenience of printing. 



38 Graphical Calculus. 

Thus though dy % standing by itself and considered apart from 
anything else, is numerically absolutely meaningless, yet when 
we write, as in the curve above, dy = 2xdx> we really mean 

that -^ = 2X, which indicates that if x increases by a small 
dx 

quantity, y increases by a quantity 2x times as great 

So in general we may write — 

8y = -/ X 8# . . . (a) 
dx 

where §>', Ex are corresponding small increments of y and x, 

which may be of definite magnitude. If, however, Sx and Sy 

are of small but finite magnitude, the equation (a) becomes an 

approximation, though usually an extremely close one, and 

not an absolute equality. 

dy 
Sx is here called a " differential," and — therefore a 

dx 

"differential coefficient" (see illustration, § 25). 

The fact that two quantities of indefinite magnitude can have 

a definite ratio sometimes causes students trouble. This may 

be got over by reflecting that the quantities 271 and $n have 

always the ratio f whatever the absolute value of 2n or 3//* 



§ 19- 

It is easy to see that when y = constant, i.e. when the 

primary curve is a straight line parallel to OX, since the 

dy 
slope is at all points of this line nothing, -— = o. Or, regarded 

dx 

algebraically, the statement y = constant is equivalent to the 
statement that y does not vary, and therefore dy (which means 
the amount of variation of y corresponding to a variation dx 
in x) = o, and therefore — 

£.= «, 

dx 



Nomenclature and General Principles. 39 

§ 20. Successive Differentiation. 

At § 17 it was explained how it was sometimes necessary 

to differentiate a derived curve. The primary curve in that 

case was a time-distance curve, the first derived a time-velocity, 

and the second derived a time-acceleration curve. Now, the 

height of the primary being denoted by y, and that of the first 

dy 
derived by y or — , the height of the second derived may, 
dx 

doc 
on the same principle, be denoted by y" or — — The latter, 

dx 

however, is very inconvenient to w r rite and print. It is 

therefore shortened by treating it as a simple fraction in which 

d stands for some definite algebraical quantity. (In reality, 

of course, it does not mean anything of the kind.) Thus we 

have — 



Xdx) 



d*y 
dx (dx) 2 

d 2 y 

or, omitting the bracket, — 

ax* 

The student must be careful to notice that this quasi- 

fraction is nothing but a shortened form of _f , that the 

dx 

expression has absolutely nothing to do with x°% and that 

at present he may regard d 2 as merely symbolical. In the 

same way, the height of the third derived is — 



d 2 y 
dx 2 
dx dx s 



d dx 2 a 3 y 



#y 
that of the fourth, — ■ and so on. 

dx x 



40 Graphical Calculus. 

§ 21. Notation of Integration. 

It will be seen from § 13 that the process of graphical 
integration consists of a construction whereby a curve is 
obtained of which the tangent of angle of slope is at all points 
= ordinate of curve we wish to integrate. The algebraical 
process is the exact counterpart of this, and consists in 
obtaining an expression which, when differentiated algebraically, 
will give as a result the expression which we wish to integrate. 
There is no general method of performing this reverse operation. 
Indeed, in a great number of cases it cannot be performed 
at all except by the aid of an infinite series. We are in all 
cases obliged to rely on our previous experience of differentia- 
tion. If the expression is of a type of which we have had no 
previous experience, we cannot do anything with it until we 
have twisted it into a shape which we do recognize as the 
result of some differentiation with which we are already 
acquainted. 

Suppose, for instance, we wish to integrate $x 2 . This 
means thaty = $x 2 is to be the first derived of the curve we 
wish to find. The problem is stated thus for an " indefinite " 
integral — 

J*$x 2 dx 

or, in the case of a definite integral — 

Co 

$x 2 dx 
a 
These expressions will be presently explained. 

This symbol f may be regarded in two ways : (1) It may 
be taken simply as a question mark. The meaning then is — 

j $x 2 d x 

? expression will give $x 2 when differentiated with respect to x 
(2) It may be taken to be the letter s, the first letter of the 
word " sum," thus — 



Nomenclature and General Principles. 



4i 




$X 2 dx 

a 
The sum between the ordinates {x-b) and (x = a) of all such rectangles as ^x 2 dx 

For it is evident that the area in square inches of a very thin 

vertical strip of the curve such as that Y 

shown in Fig. 21 is $x 2 dx (see §§ 13 

and 17), and the sum of all the thin 

strips into which the area between b and 

a may be divided = whole area of curve 

between these two ordinates = difference 

between corresponding ordinates of upper 

curve, as already explained (§ 13). 

Now, let us consider what expression Fig. 21. 

will give 3^ 2 when differentiating with respect to x as 
independent variable. (The student will understand the 
last expression better after reading the next chapter.) 

Consider what is the rule just proved (§17) for differentiating 
x n . We have found the differential coefficient to be nx ln ~ 1] . 
Hence the answer to the question fnx {n ~ 1] dx is x n . It is 
easily seen that the given expression $x 2 is of the form nx {n ~ 1] 
where n = 3, hence — 

J $x 2 dx = x 3 

A more complete solution, as will be presently explained, is 
« & 4. S ome constant." 

Hence we see that f$x*dx = x B is exactly the same 

equation as — - = 3*% but put into another form. In just 



dx 



20 



the same way as — = 5 is the same thing as 5 X 4 = 20. 

This is sometimes symbolically expressed by saying that 
f and d " cancel one another." Thus multiplying both sides 
d(x\ 



of the equation 



dx 



= $x 2 by dx we have — 



42 Graphical Calculus. 

d(x*) = 3x 2 dx 
Now multiply byf. We obtain — 

fd(x*) = J$x 2 dx 
or, since^and */ cancel — 

x 3 = f$x 2 dx 

This is not altogether a happy analogy, for f and d 
do not cancel on the right-hand side. The idea is that if 
any quantity, A (represented here by the ordinate of the 
upper curve), be divided into a large number of parts, and 
then all the parts be added together, the quantity A is re- 
produced. 

§ 22. The "Constant" in Integration. 

The expression " x 3 + constant " is known as the " in- 
definite integral " of $x 2 . It is a general expression for the 
height of every possible primary which has y y = 3X 2 for its first 
derived. We have already seen (§§ 8, 13, etc.) that there are 
an infinite number of such curves corresponding to different 
starting-points on the line OY. If a value K (suppose) be 
assigned to the " constant," the value of x* + K at any point 
also represents absolutely the area of the curve y y = 3X 2 between 
that ordinate y y and the ordinate corresponding to the point 
where the curve y = x s + K cuts the axis of x. This point 
may be found by putting y = o in the equation and solving 
for x. Thus here — 

x = - If ~K. 

This may be generally explained as follows. Suppose we 
have any curve P*Q* (Fig. 22) of which the equation is 
y =/ y (x) (where f s (x) is a shorthand symbol for " any ex- 
pression containing x "), and suppose, having integrated it, we 
obtain a curve PQ or TK, or some other parallel curve of which 






Nomenclature and General Principles. 43 

the equation is y = f(x) + c, where f(x) is some different ex- 
pression containing x. Then, in accordance with the notation 
already explained, we have ff\x)dx = f(x) + constant. 
Then the expression f(x) + c is called the indefinite integral 




Fig. 22. 

of f\x). Let x = Oq. Then for some particular value of 
5 /(*) + * = ?Q. For some other value of c r f(x) + c = ^K. 
Since the slope at K is the same as the slope at Q, the height 
^Q r is the same in each case. Now, ^Q = area of lower 
curve between the ordinate f? y (corresponding to the point 



44 Graphical Calculus. 

where the upper curve cuts OX) and </Q y ; also ^K = area 
between / r P and ^Q\ and so on. 

If we wish to find the (shaded) area of the lower curve 
between two definite ordinates, x = i and x = 2 (suppose), 
the expression for the area \sf' 2 1 f\x)dx. 

This is called a " definite integral," and we have already 
seen (§ 14) that the area is found by taking the difference in 
length between the corresponding ordinates of the upper curve. 
These ordinates are found by substituting 1 and 2 in turn for 
x in the expression/^) + c. The notation for this is/(i) + c 
and 7(2) + c, denoting respectively the ordinates a A and £B. 

Hence clearly — 



/ 



2 f\x)dx = \f(2)+c\-\f(x)+c} 

I 



= /(*)-/( I) 

Here again we see that the actual value of c is unimportant, 
since it disappears in the final result. 

The following notation is usual as a shortened form of the 
expression on the right-hand side of the above equation :— 



/ 



2 f'{x)dx= 2 \f{x) + c\ 
1 iL J 



A further exposition of these facts, with actual numerical 
examples, will be found in Chapter IX. 

The nomenclature adopted when it is necessary to 
differentiate any expression twice has been explained in § 20. 
It is also often necessary to integrate an expression twice. 
The notation then adopted is for an indefinite integral — 

ff{f"(x)}dxdx 

which means — 

fUif{*)} d *y* 

Suppose the quantity inside the [ ] brackets, viz. — 






Nomenclature and General Principles. 45 

the above expression obviously means — 
f{f(x) + c}dx 

It is necessary to notice that the constant c, which is re- 
quired (as already explained) for the first integral, must be 
included under the sign of integration for the second operation. 
The geometrical meaning of this statement will be apparent to 
any one who differentiates a curve twice and then integrates 
the result twice. Obviously, the final result will be very much 
affected by the (arbitrary) height of the starting-point for the 
first integrated curve above 0\ for the height of the curve 
obtained by the first integration determines the slope of the 
curve obtained by the second integration. 

For a definite integral the notation is — 

x c 
\f\xj\dxdx 



ra re 
J bj a 



d 
which, as before, means — 

ra r re "1 

Further explanation is not possible at this stage. 

Examples. 

1. Differentiate x 2 , x z , x 4 , x 5 , x ls , x l00 9 x a , x s , with respect to x. 

2. Differentiate {a 2 f, (b c ) 12 , {c k ) n , with respect to— 

(i.) a 2 , & c , (* 9 respectively, 
(ii.) a 9 b 9 c 9 respectively. 
(iii.) a p , b Q , c r , respectively. 

(Substitute in each case x for the quantity with respect to which the 
function is to be differentiated. Then differentiate with respect to r, and 
substitute again. Thus, in case iii. — ■ 

m 12c 

(&°)U = ff 3 * = (^) 9 = XI 
/12c 

.d\x q ) \2c W' 1 ) 12c ( :£ 7 _ ) 12c ,(&c-qA 

and = x = — x = — b ) 

dx q q 9 

3. Integrate 7r 6 , 12.1 11 , ax a ~\ 1. 



CHAPTER IV. 

general principles. 

§ 23. Changing the Independent Variable. 

We have hitherto regarded the value of the quantity denoted 
by y as absolutely dependent on the value we give to the 
independent variable denoted by x. Thus in our illustrative 
curve y = x 1 we calculated the points on it by giving arbi- 
trary values to x, and calculating the corresponding value of 
y. We have also (and we shall continue to do so) always 
plotted the independent variable horizontally. 

dy 1 
To show — = ™. — Now, we might write the relation 
dx dx 

Ty 

y - x 2 in the form x = ± \! y, and, in calculating points on 
the curve, give arbitrary values to y, and calculate x by 
finding the square roots of these values. Now, if we con- 
tinued to plot y vertically and x horizontally, we should, by 
this process, obtain exactly the same curve as before. But 
by our convention we are to plot the values of the inde- 
pendent variable (which is now y) horizontally. 

What, then, is the relation between the two curves thus 
obtained ? 

Let OPQ (Fig. 23) represent the curve y = x 2 . Let 
the curve be drawn on a piece of tracing-paper held over the 
original curve. Holding the point O fixed, turn the tracing- 
paper through a right angle in the direction of the arrow ; we 
shall thus obtain the curve OPiQi, which is clearly the same 



General Principles. 



47 



curve as before, but in which the original positive values of 
y, such as ^Q, are plotted horizontally in the negative direction, 
i.e. towards the left, as at O^. If we " reflect" the curve 
OPiQi along OY, i.e. imagine the curve turned about OY as 
axis into the position OP 2 Q 2 , this defect is remedied. This 
latter curve is the same as would have been obtained by 
plotting the curve x = \/y in the ordinary way, but with y 

Y 




p: q*p (j 



Fig. 23. 



horizontally and x vertically. If we now differentiate this 

curve graphically in the ordinary way, what we shall obtain 

dx 
will be a curve showing the values of — for all values of y 

Now, what we have to prove is that, taking any point P on 
the first curve, and the same point P 2 on the other curve — 

The tangent of slope of curve OP at point P X the tangent 
of slope of curve OP 2 at point P 2 = 1. 

It will, perhaps, appear that a simpler and more direct method of 

exhibiting this relation would have been to find the slope of the original 

curve relative to the axis of y, and to treat the axis of Y exactly as we 

dx 
have previously treated the axis of X, plotting values of — horizontally 

dy 

along a base on the left of the original axis of Y. This would, however, 
have involved temporarily reversing our ideas of positive and negative direc- 
tions. It is easy to see that the present construction is in reality the same 
as this would have been if we had also rectified the signs. The additional 
utility of the present construction will be more apparent at a later stage. 

Taking an adjacent point Q, it is clear that wherever Q is — ■ 
PM = MiQi = M 2 Q 2 and MQ = P^ = P 2 M a 



48 Graphical Calculus. 

Hence — 

MQ M 2 Q 2 = 
PM ' P 2 M 2 

And as this is true wherever Q is, it is true in the limit, when 

dy dx 
it moves up to and ultimately coincides with P, i.e. -~ X -- = i, 1 

a result which, though it was certainly to be expected, we 
should not have been justified in assuming merely because 

dv dx 

— and — look like fractions. Thus we see that, whenever 
dx dy 

dy . 
we wish to find a value for -j- in any given case, if it happens 

dx 
to be easier to find — from something we already know, we 

dy 
are at liberty to do so, and thence deduce — by inverting the 

dx 

value so found. 

Some students find a difficulty here which is not easy to express in words. 
It is as follows. This proof, as it stands, only holds when Q is at the same 
distance from P as Q 2 is from P 2 . When this is the case, it is clear that 

— X — = i. But the values of dy and dx are not definite. We only know 
dx dy 

that they are indefinitely small, and, provided they are indefinitely small, 

they can have any order of smallness. How, then, are we to know that the 

dy and dx in the first factor are the same dy and dx as in the second factor ? 

The answer is, that, provided dy and dx are indefinitely small, the value of 

dy 
the ratio — is constant for any particular point, no matter what the order 
dx 

of smallness of dy and dx. The dy and dx of the first factor must be taken 

together ; likewise the dy and dx of the second factor. The difficulty arises 

dy 
from thinking of — as a variable fraction instead of a fixed ratio of two 
dx 

variable but mutually dependent quantities. 

1 It is easy to see, from the note on p. 47, that this result is merely 

another form of the trigonometrical relation tan 9 X tanf q) = 1, where 

6 is the angle of slope of a curve at any point. 



General Principles. 49 

Thus we see that if we differentiate curve OPQ and 
OP0Q0 with OX as base, and take any two corresponding 
ordinates of the two derived curves (e.g. pV and p 2 Y 2 are cor- 
responding ordinates), the rectangle formed with these two 
derived ordinates as sides will contain exactly one square 
inch. 

Now, we have hitherto, for the sake of clearness, regarded the value of 
y as being dependent on the value of x. For the purposes of the calculus 
this is not in the least necessary. Our results are just as good if, as a 
matter of fact, the value of x depends on that of y, or if the value of 
both x and y depend on that of another variable z, which, although its 
variations may be the prime cause of the simultaneous variations of x and 
y, does not appear in the equations at all. What the calculus is really 
concerned with is the fact that x and y do actually vary together in such a 
way that every definite value of x corresponds to a definite value or values 
of y, and it is not concerned with what may or may not have been the 
ultimate cause of those simultaneous variations. 



§ 24. Differentiation of x v \ 

We can utilize this principle at once to further our proof 

dy 
of the general proposition that if y = x n i — = nx {n ~ l K 

Let n be of form — where m is an integer. 

m ° 

I i — 

Let y = x m where x m means Vx (see chapter on indices 
and surds in any algebra). 

As far as we have hitherto gone, we cannot differentiate 
this. 

It is easy to see that the curve y - x m is the same as the 
curve x = y m , for the same values of x and y satisfy both 
equations. 

If we differentiate x = y m with respect to >', we get — 



dx 
dy 



_ = my {m - 1] 



50 G rap J deal Calculus. 

This operation corresponds exactly to turning our curve through a 
right angle, reflecting along OY, and differentiating graphically. 

Hence we have — 

dy i 

dx my {m ~ J) 

dy 
But we require the value of — in terms of x, and not of y. 

dx 

Substituting x m for y, we get — 
dy i i 



dx m ( iV*- 1 



\X" 

I I 

x m 

T (i-l) 

m 



which is of the required form. Thus the proposition holds good 



when 7t is of the form --. 

771 



§ 25. Illustrative Example. 

The following example will give an idea of the practical 
meaning of this principle. 

Suppose we take a barometer up a mountain-side from sea- 
level, and note the height of the barometer at short intervals of 
vertical rise. (The vertical heights must, of course, be known, in- 
dependently of the barometer.) Let a height-barometer reading 
curve be plotted from these observations. Let 100 feet = 1 unit 
on the diagram, and barometer-reading scale be full size. The 
characteristics of the curve that would be obtained by such 
a proceeding are shown exaggerated in Fig. 24. On differen- 
tiating this curve graphically, we shall obtain a first derived, 
which, since the primary always slopes downwards, lies 
entirely below O r X\ Thus p y ? y shows the rate of rise i or, 



General Priitczples. 



5i 



in other, words, P^ shows the rate of fall , of the barometer 
at P per 100 feet lift. 

Notice carefully the significance of the signs here. A fall is a negative 
rise. If the barometer falls + 0*5 inch, it may be said to rise — 0*5 inch. 

Thus, suppose /P^ 0*5 inch. The meaning of this is that 



at the point P the rate of fall is 



0.5 inch 



100 feet lift 



— (see note on p. 30). 



29"- 


4^ 


p 
















V 










^ 










1 








28"- 


> 




| 


eetO 


^ 50 ° 


p 


600 




'3 
















k. 








of 








% 








^ 










P 






^ 








§ 








Ss 


















D N 






<X 









Fig. 24. 



Now, we may, as at § 18, p. 38, conveniently write o> = — 

dx 
X $x ; e.g. assume that we lift the instrument through 6 inches 
(8x), when at an altitude given by Op. The instrument will 



dy 



dy 



rise (see note above) by an amount 8v =— x Sx where — 

dx ~ dx 
^ f^ s is the current rate of rise per 100 feet lift Hence 

0*5 inch 

By = - ^-7— x °"5 fo °t 
100 foot 



5 2 Graphical Calculus. 

The dimension " foot " cancels out 

0*5 inch 

= — ■ = — x-^o lnc h 

200 * uo 

that is to say, the barometer falls -^ of an inch. 

If the student be not well versed in the method of 

dimensions, he should carefully note the illustration given here. 

Of course, this only holds where Sx is so small that the point 

on the curve whose abscissa is (x + Sx) is not any appreciable 

distance from the tangent to the curve at the point P. It 

8 v dy Sy dy 

implies that — = — . If - differs sensibly from — , there will 
ox ax ox dx 

be an error introduced. This point has been already explained 

several times (§§ 7, 12, 13, 15, etc.) in various aspects. If the 

student does not understand it, he is referred to the sections 

quoted. 

Now, suppose we turn our primary curve through a right 

angle and reflect it on OY. We shall obtain the upper curve 

in Fig. 25, which is exactly the same curve as Fig. 24 viewed 

under another aspect. The height of the derived curve now 

represents the vertical distance through which we must lift 

the barometer in order that it may fall 1 inch, assuming that the 

rate of vertical lift per inch fall of the barometer remains 

constant ; or, in other words, p y F x represents the instantaneous 

rate of lifting per inch rise of barometer. Now, what we have 

just proved amounts in this case to this — 

Rate of vertical lift (inhundreds of feet) per inch fall of barometer 

1 

rate of fall of barometer in inches per 100 feet lift 

In the particular case considered above, rate of fall of 
barometer per 100 feet lifted through was 0*5 inch per 100 
feet at a certain point. Hence, at that point rate of lifting 

per 1 inch fall of barometer = — = 2 : i.e. the instrument 
r 0-5 



General Principles, 



53 



must be lifted 2 units or 200 feet if the barometer is to fall 
1 inch (assuming constant rate of falling), which is, of course, 
otherwise obvious. 

This illustration may probably present some difficulty to 
the student, partly owing to the essential difficulty (to a 
beginner) of viewing the same ratio under two aspects, and 
partly from the confusion introduced by the practically 



v 

]\ 

0~ p 

I 
I 
I 
I 
I 

! 

& !£_ 

Y' / 



Fig. 25. 

necessary difference of scale in the vertical and horizontal 
directions. We have already had several instances in which 
the scale was intentionally arranged so as to be as simple 
as possible. This is intended as an exercise in variation of 
scale. The best way to understand confusing examples of 
this kind is to keep the mind fixed on the diagram rather than 
on the form of the words. 



54 Graphical Calculus. 

Exercise. — Draw a curve of any shape and differentiate it. 
Turn it through a right angle, as explained in § 23. Reflect 
on OY, and differentiate again. Mark corresponding points 
on the two curves, and show by the method explained in § 2, 
Fig. 3, that the mean proportional between the heights of the 
derived curves is always 1 inch. (This is, of course, merely an 
application of the principle that tan 6 X cot 6 = 1.) 

§ 26. Differentiation of Sum and Difference 
of Functions. 

Suppose we have given two elementary curves; for 
example, (T) and (2} in Fig. 26, which represent y = x 2 anc 
y = fj^ Draw another curve whose ordinates are = the I 
sums of the corresponding ordinates of the given curves I 
This may easily be done graphically. In Fig. 26, all pairs o 
corresponding ordinates (such as P/, ^Q) of and 
are together equal to the ordinate (such as rR) of 0. I 
Differentiate (7} and (0 and place the derived curves on the 
immediate right of the corresponding primaries ; thus (4} ) 
is the first derived of (T), (7\ of 0, and (6\ of 0. We \ 
shall now show that there is the same relation between the 
ordinates of (0 (7), and (0 as there is between those of J 
Q, Q, and 0, e.g. that/P + $ y Q = /R\ 

Froof. — By construction — 

f? + ^R = rR 
that is, i-L + /M = uN ; 

also by construction — 

Hence by subtraction — 

LS + MT - NU 



General Principles. 

Dividing through by PL = QM = RN, we have- 

LS MT NU 

FL QM ~ RN 



55 




Fig. 26. 



that is, when S is close to P, and therefore T and U close 
to Q and R respectively — 

/P y + $Q = r y R" 

(see note on p. 48). 






56 



Graphical Calculus. 



Roughly speaking, the meaning of this is, that if we have two slopes 
(which we may imagine as wedges cut out of a pack of cards) of the same 
length piled on top of one another in the way shown in Fig. 27, the 
resulting slope (tangent of angle) is the same as that of the other two added 
together. This can be easily seen. 

Now, the equation to curve (7) in Fig. 26 is evidently — 



Ord.of( 3 ). 

y 



Ord. of(i). 



+ 



Ord. of (2). 



Our result tells us that its derived equation is- 



Slope of (3). 



Slope of (1). 
2X 



-f 



Slope of (2). 



as already shown in §§ 17 and 24. 

Exactly similar reasoning applies to the differential 
co-efficient of differences of func- 
tions. In this case the ordinates 
of curve (7) are to be made = the 
differences of the ordinates of (T) 
and (2^. The figure can be easily 
made from Fig. 26, by exchanging 
the places occupied by curves (T) 
and (7), and also those of (4) and 
(6^ It is then easily seen that the 
former proof applies also to this. 
Indeed, the proof given above for 
the sum will hold throughout in- 
dependently if we change the + sign 
into — . If the wedges (1) and (3) 
in Fig. 27 also change places, the result may be seen to be 
the same as the + result viewed under another aspect. 
Generally speaking, our result may be written thus : 




Fig. 27. 



Ii y = u + v — w + t — r, etc., where 
functions of x, we have — 



/, v, etc., are any 









General Principles. 57 

dy dn dv dw dt dr 
dx dx dx dx dx dx 

There is no difficulty in extending the proof in this 
manner. It is left as an exercise for the student. 



§ 27. Illustrative Example. 

A simple practical example of this principle, which, though 
not scientifically quite accurate, as will presently be explained, 
is instructive and easy to understand, is as follows. 

A man holds two appointments, in one of which his salary 
is ;£i8o, with an annual increase of ^"20 per annum. In 
the other his salary is ^85, with an annual increase of ^15 
per annum. His income tax at this time is ^7 per annum, 
and is increasing at the rate of jQ\ per year per year. It 
is required to find what is the total net rate of increase of 
income. 

Call the salaries y u y 2 , the income tax y 3 , and total 
income y. Let x represent the number of years reckoned 
from this time. We have y = y x + y 2 — y B . It is clear that, 
if we take Ax to be any integral number of years — 

— = rate of increase on first appointment 

Ax 



Ay 2 = 

Ax 
Ax 



second 



income tax 



. . Ay Ay, Ai'o 

In this simple case it is easy to see that — - = ~- + -^ 

Ax Ax Ax 

— , which is another way of saying that total net increase 

Ax 

of income = rate of increase of salary on first appointment + 
rate of increase of salary on second appointment — rate of 
increase of income tax. 

This notation is not strictly applicable to the case in point, 



58 



Graphical Calculus. 



because the rate of pay does not increase every instant, but 
step by step, each step being one year broad, and we have, 

B therefore, to assume such 
particular values for Ax as 
will make the expressions 



A 



u^ 



„S_ 




Years J 2 

Fig. 28. 



dx 



— , etc., give a correct idea 
Ax 

of the rate of increase. If 

the salary or rate of pay 

increased every instant, this 

stepped figure (28) would 

merge into a straight line 

as shown, and the notation 



could then have been applied to it. 



Exercise. — The line AB, Fig. 28, is itself the derived curve 
of another curve. What do the ordinates of this other curve 
represent ? Ans. The aggregate earnings of the man. 

Another illustration of a more scientific character will be 
found in the following. A man in a corridor train commences 
to walk along the corridor in the same direction as the train 
is moving. Suppose the ordinate of curve (J), Fig. 26, repre- 
sents the distance travelled by the train in a time (after the 
moment of starting) represented by the abscissa. Let curve 
Q represent in the same way the distance walked by the 
man along the corridor (or, as it is expressed in kinematics, 
" relative to the train "). Then curve (7) represents the total 
distance moved by the man through space (relatively to the 
ground) in time represented by the corresponding abscissa. 
Our principle states that ordinate of derived curve of (T) 
+ ordinate of derived curve of (T) = ordinate of derived curve 
of C$\i which, as we see from § 17, is equivalent to stating 
that at any instant velocity of train + velocity of man along 



General Principles. 59 

corridor = total velocity of man relative to rails. Differentiating 
again, we have acceleration of train + acceleration of man 
along corridor = acceleration of man relative to rails. 

Let the student follow out the case where the man walks 
in the opposite direction along the corridor. 



§ 28. To Differentiate n/(x). 

An important analytical principle can be deduced from 
a special case of this result. 

Suppose each of the two curves (7) and (7) in Fig. 26 
had been exactly alike ; then curve (7) would be twice as 
high as either Q or (7), and (7) would be twice as high as 
(a) or (Jj* Thus, when u is any function of x — 

if y = 211 

dy du 

then -f = 2 — 

ax ax 

dy dv 

or ity = 3^— = 3~ 

dx dx 

This law evidently holds for any integer whatsoever. It 
also clearly holds for any fraction. For in this case curve 
Q is half as high as curve (T), so that the derived of Q 
is half as high as the derived of (7). Proceeding in this way, 

we can prove the principle for any positive or negative integer 
or fraction. 

Hence, when n is any quantity — 

dy du 

\i y = nu, — = n — 

dx dx 

Exercise. —Prove the result when n is a negative quantity. 



60 Graphical Calculus. 

§ 29. Integrals of Sums and Differences of 
Functions. 

It is obvious that the principle proved in § 26 applies 
equally well to integrations, for in Fig. 26 the curves (T), 
(T), and (J^ are respectively the integrals of (4), (7), and (T). 
Now, Q is the sum of (4) and (7), and the ordinate of (7) 
represents its area. The ordinate of (?) represents the area 
of (7) (reckoned as explained at p. 42), and that of (7) the 
area of (4^. Hence since (T) + (T) = (7), it is clear that — 

Area of (4) + area of (7) = area of (6^ 

This proof assumes as self-evident the fact that a curve cannot have 
more than one first derived curve. 

This may easily be proved independently, for if we take a 
corresponding " element of area n of each curve (as an 
infinitely thin strip is called), such as that shaded in the 
figure, it is clear that since — 

fT + ?V = r y BS 

area of strip of (4) + area of strip of (7) = area of strip of (6^ 

The same relation holds for all the common strips into 
which each curve may be divided. It therefore holds for 
the sum of all the strips, ue. for the whole areas of the curves. 
The same is obviously true mutatis mutandis for the difference 
of two curves. 

As a special case of this, we see that — 

Jnf(x)dx = nff(x)dx 

where n is any quantity whatever, and f(x) has the meaning 
already explained on p. 42. 



General Principles. 61 

For since nj{x) =/(x) +/(x) + . . . to n terms, we have 
as above fnf(x)dx = f f(x)dx + f f(x)dx ... to ^ terms 
= n ff(x)dx. 

We can now easily find the integral of any multiple of any 

power of x. Suppose we require f $x 5 dx. The power of 

the integral must clearly be x G . Let us differentiate x Q , and 

compare the result with the proposed expression. We obtain 

6x*. This is clearly twice too great, so the desired integral 

is \x Q . 

P 
Similarly, required f -j^x r dx (where /, q, r represent 

either numerical quantities or expressions whose values 
do not depend on that of x), we find that the d.c. of 

P 
x r+1 is (r + i)x r . This must be multiplied by , 

to obtain the desired result ; hence the required integral is 
P 



-x 



.(r + l) 



*lq{r -f- i) 

Many other expressions can be reduced by simple alge- 
braical or trigonometrical operations to forms which can be 
differentiated or integrated by means of these rules. 

For instance, to find/*^ — a) 2 dx, we have — 

(x — a) 2 = x 2 — 2ax + a 2 y and therefore 
f(x — cifdx =fx 2 dx — 2cifxdx +fa 2 dx 

= ax' + arx 

3 

Examples. 

1. Differentiate (by expanding) {x + 3) 2 , (x -f a)\ (i + i6x 2 + 64* 4 )*, 
(x + a){x - b). Am. 2{x + 3), 4{x + a) 3 , i6x, 2x + a-b. 

2. Integrate — 

(i.) ( "^L' X -^ ) . An, ,— r^— Tx /j> 
V \cjd V/ / (p + q-\-i)<:\/ df 



62 Graphical Calculus. 



(ii.) (x + 3) 3 . Ans. - + 3.r 3 + ^x 1 + 27*. 

(iii.) (x — 3) 3 . ^;/j-. 3-* 3 -1 — -x 2 —27^. 

4 2 

jc 3 

(iv. ) (x + a) (x - a). Ans. a"x. 

(V.) \! X 8 (I + 2rt 4 + a 8 ). ^*J. ^-^^ 5 . 






CHAPTER V. 1 

GENERAL PRINCIPLES {continued)* 

§ 30. Products of Functions. 

It must be carefully noticed that the principles explained in 
the last chapter cannot be extended by analogy to multi- 
plication and division of functions. 

For instance, if — 

y = uv 

where uv stand for any expressions involving x, it by no means 
follows that — 

dy dii dv 



dx 



or that if — 



that therefore- 



dx 


■ dx 


y = 


U 

V 


dx 


du 
dx 
dv 




dx 



1 In this and the succeeding chapters almost ail the diagrams are 
reduced copies of drawings made to scale. In many cases the scale with 
which the curves are to be measured is given. The student should always 
measure the curves, and make as large accurate drawings to scale for 
himself as possible. Three times full size is a convenient scale for a half- 
imperial sheet. 



64 Graphical Calculus. 

The student must be very much on his guard against 

assuming results from analogies of this kind. He should in 

all cases return to the curves, and think each principle out 

on its own merits. 

dluv) 
The graphical proof of the formula for — — will be best 

ax 

understood if we first give a brief algebraical one. 

Suppose we have three variable and mutually related 
quantities, which we shall denote (i), (2), (3) respectively, 
of which the values of (1) and (2) both depend directly on 
the value of an independent variable x, so that curves can be 
obtained which show the relations between (1) and x and 
between (2) and x. When x has any value we like to give it, 
(1) and (2) each assume definite corresponding values. 

Now (3) is to vary in such a way that, whatever the value 
of x, its value is always equal to the product of the corre- 
sponding values of (1) and (2). Thus it is clear that the value 
of (3) must also depend entirely on the value of x, and the 
problem is — if the value of x changes slightly at a time when 
the values of (1), (2), and (3) are respectively u, v y and y — to 
find what is the relative magnitude of the consequent change 
in the value of (3). 

Obviously, from the data — 

y = uv . . . (i.) 

Now, if x changes from the value it now has, it is clear 
that the values of (1), (2), and (3) will also change from the 
values u, v, and y respectively. 

Suppose a change in the value of x of the magnitude Ax 
causes the three quantities to become (u + A//), (v + Av), 
(y + Aj), respectively. Then, since (3) always = (1) x (2), we 
must have — 

y + Ay = (u + An) [v + Av) 

On multiplying out, this becomes — 

y + Ay = uv + uAv + vAu + Az/Az> 



General Principles. 65 

but since j = uv, we have, on subtraction — 

Ay = u Av + vAu + A#A# . . (ii.) 
and therefore — 

Ay Az; , Au . Av . ,... * 

Ax Ax Ax ' A.t 

If this change in x had been infinitely small, all the 
quantities, Ay, Au, Av, would have been infinitely small too, 
and the equation (iii.) would have been — 

dy dv du , dv ' ,. . 

— = u— + v- + —du . . (iv.) 
dx dx dx dx 

dy dv du . 

Now, — , — -, and — are all of finite magnitude, although 
dx dx dx 

dy, du, dv, dx are infinitely small (as already explained in 

§§ 5, 17, 18, etc.). Hence the last term in (iv.), being the 

product of a finite quantity with an infinitely small one, does 

not affect the equation at all, as it is infinitely small compared 

to the other terms (see § 12). Therefore we have — 

dy dv du 
dx dx dx 

That is to say, when the variable quantities (1), (2), (3) 
have the values u, v, y, the rate of change of (3) per unit 
increase of x - u x rate of change of (2) + v X rate of change 
of(i). 

This result may be very clearly demonstrated graphically. 

The ordinates of curves (1), (2), (3) (Fig. 29) represent the 
values of the three variables for all values of x. The length 
of ordinate of (3) always = product of corresponding ordinates 
of (1) and (2). Curves (1) and (2) being given, and the scale 
with which they are to be measured, (3) can always be found. 
Thus if pV = o*4, and ^Q = 2*0, then rR = o*8, and 
so on. 

F 



66 



Graphical Calculus. 



Let/P, ^Q, ;-R be the definite values u, v,y respectively, 
and let PL = QM = RN be the value Ax. 




Fig. 29. 

Then LS = A21, MT = Av, NU = Ay 
Draw a rectangle ACDB, of which AB =/P, AC = ^Q. 



General Principles. 6j 

It is clear that.the number of square inches (see note on p. 17) 
in this rectangle = number of inches in rR. Produce AB, 
AC, so that AE = *S, AF = fT. Then- 
Rectangle AFGE = u\3 
hence gnomon EDF == NU = Ay 
But gnomon EDF = rect. ED + rect. DF + rect. DG 
= BD . BE + CD . CF + CF . BE 
i.e. NU = Ay = j 2 A^ +jAy% + AftAft 

All these increments have been produced by an increment 
PL = Ax in x. 

This is true however near Q is to P, or however far off 
it is. It is therefore true when PL is infinitely small. But 
when PL is infinitely small, the small rectangle DG is infinitely 
small compared to the rectangles ED, DF; for, comparing 
the area of rectangle DG with that of, say, FD, when PL and 
therefore also CF and BE have become infinitely small, we 
see that, although each of these rectangles has the same 
breadth, CF, yet the length of DF, viz. CD, being of finite 
magnitude, contains an infinite number of lines = EB, which 
is the length of the rectangle DG, and therefore rectangle DG 
is infinitely smaller than DF. 

Thus Ay 1 Ay 2 vanishes in the equation for Ay, being infinitely 
small compared to the other terms, and our equation may be 
written dy — udv + vdu (see pp. 37 and 21), remembering that 
all these increments have been produced by an increment dx 
in x. This may be signified to the eye by writing equation 
in the form — 

dy dv du 

dx dx dx 

To illustrate this, differentiate curves (1) and (2), placing 
each derived curve on the immediate right of its primary at (4) 
and (5). Then multiply curves (1) and (5) together, just as 
(1) and (2) were multiplied together to produce (3), 1 and place 
the result at (7). Do the same with (2) and (4), placing 
1 See note (6), p. 194. 



68 Graphical Calculus. 

the resulting curve at (6). Now, on adding together (6) and 
(7), we shall, as the proof shows, obtain (8), which is the first 
derived of (3), as may be found by trial. 

The actual equations to the curves are shown in the 
diagram. 



§ 31. Illustration of D.C. of Product. 

Before proceeding to a more general case of products, 
we shall consider an example of the application of this 
principle. Given that at the present time the total number 
of poor-houses in the country is 7251, and the average number 
of paupers in each 112, the rate of increase in the number of 
poor-houses in the country is 8 per year, and the annual 
decrease of the average number of paupers in each is 17. 
Find how the total number of paupers in the country is 
varying. 

It is clear that — 

Total number of paupers = total number of poor-houses X 

average number of persons in 
each. 
Let y = total number of paupers. 
y\ = total number of poor-houses. 
y 2 = average number of paupers in each. 

Then y = y\ x y 2 

Let x be the number of years reckoned from the present 
time. 

Now — 

dy\ . 

— = rate of increase of poor-houses 

ax 

= +8 

— = rate of increase of average persons in each 

ax 

= -i'7 



General Principles. 69 

dy 

S- ~ required rate of increase of total number of 

dx 

paupers. 

Henoe we have, from what we have just proved — 

dy 

-f- = -7251 X 17 + 112x8 
dx 

= -ii,43°7- 

Hence, from these data, the total number of paupers is 
decreasing at the rate of 11,4307 annually. 

Exercise. — Why is this number not exactly the same as 

that which would have been obtained by finding the value of 

(7251 x 112 - 7259 x 110-3)? 

dy 
Ans. Because the rate of increase — is not constant. 

ax 

The value here found for it is only the momentary rate of 

increase (see remark on p. 14). 



§ 32. D.C. of a Continued Product. 

We can easily proceed from this result to a more general 
expression for the continued product of a number of functions. 
Suppose, for instance, y - uvw> where u y v, and w are, as before, 
shorthand symbols for " any expressions involving x" 

Here our primary curves are y l = u, y 2 = v, y 3 - u>, etc. 

The above equation may be written as a product of two 
quantities, thus — 

y = (iiv) X w 

In this form we can differentiate it by the previous section, 
thus — 

dy d(uv) , dw 

-+- - w— — - + uv — 
dx dx dx 

We can differentiate (uv) as before, so as to get the first term 
cf the right-hand side in a simpler form — 



yo Graphical Calculus. 



dy 


r du . dv \ . dw 

W V (- u— 1 + uv — 

\ dx dx ' dx 


dx 



Removing the bracket, we obtain — 

dy dw , du , dv 

-z- = uv f- vw h wu — 

dx dx dx dx 

In the same way we can proceed to the differentiation of 
the continued product of four functions, the result being 
as follows — 

If y = r stu- 
dy du i dr , ds dt 

— = rst— — h siu h tur — + urs — 

dx dx dx dx dx 

These results are often more conveniently and sym- 
metrically written — 

i dy i du , i dv , i ds 
-f-.= - — +-— +--, etc. 
y dx u dx v dx s dx 

the result being obtained by dividing each side of the upper 
equation by the corresponding side of the equation y = urst. 

Exercise, — Draw three curves at random, and a fourth 
showing the value of the product of the three. Differentiate 
them graphically, and exhibit the truth of the above result as 
accurately as possible. Prove the result independently of the 
proof in § 30. 

The chief difficulty in understanding this result is due to the multitude 
of different symbols which are often (as the student is prone to think) 
needlessly introduced into the proof. He is apt, for instance, to stumble 
over the symbols u, v, etc., and to ask himself, in the case of such an 
equation as y = u, " What is the use of introducing u at all, if we are 
already dealing with a quantity y } which denotes exactly the same thing ? " 
The answer to this is that, whereas y stands for the ordinate to the curve, 
u is used for brevity, instead of (fx), and means " some expression in- 
volving .r," and may stand for any such expression ; and the equation 
y = u means " there are, for any value of x, one or more definite values 
for y ;" i.e. "jrandjy are dependent on one another. " If the student finds 
other difficulties of this kind, the best plan is to express his difficulty in 



General Principles. J\ 

words and write it out. It will, in most cases, be found that the very 
exercise of explaining accurately to himself what his difficulty is (besides 
being of high educational value in itself), will enable him to explain 
the difficulty away. To obtain a clear understanding of any point, 
there is nothing like seeking for a geometrical explanation by assuming 
curves about which to reason instead of symbols. It is much easier 
to reason about the curves themselves than about the symbols denoting 
them. 



§ 2>Z* Orders of Infinitesimals. 

The case of the d.c. of the continued product of three 
functions may be proved independently of § 30, and by the 
same method as was adopted in that section. This proof 
illustrates very well the principle of what are called " orders 
of infinitesimals." While leaving the working out of the com- 
plete proof as an easy exercise for the student, we shall give 
as much of it as will enable us to show the meaning of this 
expression. 

Suppose the ordinates of curve (0 4 ), in Fig. 30, represent 
the products of the corresponding ordinates of (OJ, (0 2 ), (0 3 ). 
Imagine a rectangular block made, the lengths of whose 
edges are equal to a particular set of corresponding ordinates 
of three given curves. 

Let O l p = 2 q=, etc., be the current value of the 
independent variable. 

Let ab = pV, be = aQ, mn — rR. 

Then the number of cubic inches in the white block (of 
which only one face, abed, can be seen in this view) = the 
number of inches in $S. 

Now, if x increases by A# = // =, etc., then a conse- 
quent simultaneous increase, A^, Ay 2 , Ajk 3 , and Ay, will take 
place in y x , y 2 , y Si and y. These increments are respectively 
KT, LU, MV, NW. The block, therefore, increases to 
amzjhe. The number of cubic inches in this increment = 
NW. This increment consists of all the shaded parts of the 
block, together with a slab at the back, which in this view 



72 



Graphical Calcuhis. 



is entirely concealed. This cubical increment is made of 
several parts. 

(i.) Three large flat plates (shaded light in Fig. 30), degfc, 








T^- — ^ 






P 




K ^ 






h 




t 




o, 


1 
1 


\\ 






Q, 


^ 


L 






/ <7 




IL 




o, 


/ ' 1 
/ 11' 




/ R 




m 






y^ r 




V 




3 


1 


__— 




s/ \ 




N 






f s 




W 




°4 


J 









Fig. 30. 

cbmnlk, and the hidden plate at the back. These are repre- 
sented by y z y Ay* y%yAyu }\yAy* respectively. 

(ii.) Three rectangular four-sided prisms (shaded darker 
in the figurej, g/ii, lnz 1 fck!\ the magnitudes of these are 
respectively, yAy.Ay 3 , yAy* &j\, y s AyAy* 



General Principles. 73 

(iii.) The small rectangular piece ijl (black), whose 
magnitude is A)\Ay 2 Ay 3 . 

Now, suppose Ax, and consequently also Ay l} Ay 2 , Ay 3 , and 
Ay, to dwindle indefinitely till they are infinitely small = dx, 
dy l dy 2 , dy 3 , and dy. Then — 

(1) The flat plates, such as gedcf, become indefinitely thin 
in one direction (de) ; and although the other edges, such as 
eg, ef, are the same size as the original block, yet these flat 
plates are clearly, as regards their cubical contents, infinitely 
small compared to the block abed. 

(2) In the same way, the prismatic pieces (such as fckt) 
though in two directions (cf, kl) they are just the same size 
as the plates {gedcf), nevertheless become cubically infinitely 
small compared to the plates. 

(3) Again, the small piece (black), though in every two 
directions it is the same size as one of the darkly shaded 
prismatic pieces, is nevertheless infinitely small compared to 
any of them. 

Hence we see that though we may have any number of 
different quantities of the same kind, and all infinitely small, 
yet they may have " orders " of smallness among themselves, 
i.e. one quantity (3) may be infinitely small compared to 
another quantity (2), which is itself infinitely small compared 
to another quantity (1), which in turn may be infinitely small 
compared to a quantity y, and so on. This would be expressed 
by saying that the plates are an " infinitesimal of the first order," 
the prismatic pieces "an infinitesimal of the second order," 
and the small cubical piece " of the third order," and so on. 

We have already had several instances of infinitesimals of different 
orders. Thus in § 13, Fig. 11, what we showed with respect to each of 
the infinitely small "elementary" vertical rectangles of which the lower 
curve was composed was in reality, that each rectangle differed from the 
corresponding strip of the curve by a small triangle, which was infinitely 
small compared to the infinitely thin rectangle ; in other words, that the 
difference was an " infinitesimal of the second order," and the sum of an 
infinite number of these infinitesimals of the second order was, comparable 
in size with the rectangle K'D, an infinitesimal of the first order. 



74 Graphical Calctilus. 

§ 34. D.C. of x n . 
We are now, for the first time, in a position to complete 
the proof of the formula for d.c. of x n . 

We have not hitherto proved that it is true either for such 
p 
expressions as a» or x { ~ m \ 
p 

If y = X q 

111 

y = x q X x 1 x x 1 . . . to p factors 

We can differentiate it in this form from the rule for 

1 
continued products, for we have seen (§ 24) that when y = x 9 

the rule holds good. 

Hence we have, from §32 — 

dy I I 1 (Hi) 

— = xfl X x* X . . . to p — 1 factors X -x^ q 
dx q 

+ the same quantity (p — 1) times repeated 
since all the factors are alike 



= / x < \#*/ x-^ 7 f 

x ■ 1 






= - x q X 

/ (^-1) 
= -x^ q ' 

which is of the required form. 
Now, let us suppose that — 

y = x~ m 
where m itself, apart from its sign, is any integral or fractional 
positive quantity. 

Another way of writing the equation is — 

1 
y ~~^ 

Now, we shall find — indirectly thus. Find an expression 
dx 

for the differential coefficient of x m X — , which we know 

J x n 



General Principles. 75 



from other sources = (for x m X — = ij and from § 19 

dy 
we know that when y = 1 -— = o). 

Having found this expression, if we equate it to zero we 



shall have a simple equation involving , which by 

solution will give us what we require in terms of the d.c. of x m } 
which we know : 

( x m X —) d{ — 

x m J \x m ; , 1 



T 



Hence- 



therefore- 



dx dx x ' 



X m J I 



» X , 

x m -\ mx {m - 1] = o 

dx * m 



\ x m J xf 71 



Um - 1) 



dx 



= —mx l '~ m ~ Vl 

As this proof is usually given, it involves a difficulty to the beginner 
which he often finds difficult to express in words. It is usual to write— 

If>= — 

then yx m = I 

Differentiating both sides of this equation, we have, etc. Though the 
student cannot find anything to actively object to in the words in italics, 
and though he may understand the process of differentiating a product, 
yet, because he does not understand the meaning of the reasoning, the 
proof fails to convince him. If he compares the reasoning given above 
with what is usually given, i.e. if he substitutes for y in terms of x in 
the product to be differentiated, he will find it easier to understand. 



y6 Graphical Calculus. 

Taking the meaning of the italicized words, literally they may be 

assumed to mean that if two curves, z = yx m and 2=1, are the 

same, then their derived curves are also the same with respect to any 

variable whatever. The fact that one of the factors of the product yx m > 

viz. y, does not contain x, need not trouble us, for we know that although 

y does not appear to contain x at all, yet it does so in reality, for the 

value of y may be expressed, if we please, in terms of x. Indeed, if it did 

dy 
not depend on x the expression -j- would be utterly meaningless. 



§ 35. D.C. of a Quotient. 

We can easily find the d.c. of a quotient of two functions 

u 
of x by an application of this principle ; for suppose y = -, 

where u and v are any functions of x — 
Then yv = u 

Differentiating both sides of this equation by the product-rule 

on p. 65— 

dv dy du 

dx dx dx 
This is a simple equation to find -- giving— - 



or, substituting - for y- 



If this is not clear to the student, let him substitute, as 
an example, say, (x'r) for u, and (x$) for v, and f — J or (xk) 

for/, throughout the proof here given. 

dy 
The expression for — should be committed to memory. 



du 
dy dx 


dv 
dx 


dx 


V 


du 

v 

dx 


dv 
- u— 

dx 



General Principles. yy 

The same thing could be proved directly from the curves 
Fig. 29. The bracketed numbers refer to the ordinates of 
the corresponding curves, and d(i) means the " first derived 

°' ( "-" (3) 

Given (2) = pr, required the equation of (5). We have — 

(7) _ (8) - (6) 







«'-w- 


(0 






43)- 


(I) 






(1M3 


) - (3)*(i) 
(O 2 


Or we 


might prove the same 


thing thus : 




y = 








= 


uv' 1 






dy 
dx 


dx 


.die 

- l — 

dx 



( o dv 7 1 dn 

V ' dx J v dx 

(The student will not understand this last step till Chapter 

VIII. is reached.) 

du dv 
dx dx 



1? 

sin x dv 

As an example, we might have y = 1 to find — . We 

log x dx 

can write at once — 

, d(sin x) . d(log x) 

log x — — sin x— — - — - 

dy_ dx dx 

dx (log xf 

which we cannot as yet further simplify. 



78 Graphical Calculus. 

Every practical example of the product-rule furnishes also 
an example of this rule. 

It is to be noticed that, since in the plate ordinate of 
(6) -f ordinate of (7) = ordinate of (8), therefore — 

area of (6) + area of (7) = area of (8) 
i.e. area of (6) = area of (8) — area of (7) 
i.e. area of (6) = ordinate of (3) — area of (7) 

all areas being taken between corresponding ordinates, or 
reckoned as explained at p. 42. The bearing of this on the 
integration of expressions will be explained later on. 

Examples. 

1. From the illustrative example given in § 31, find, by inversion of 
this, the rate at which the poor-houses in the country are increasing, given 
total number of paupers (= 112 x 7251) and the average in each poor- 
house (112) and the rates of increase of these ( — 11,430 and— 17 
respectively). 

d(x 2 ) d(x z ) 

2. Find from the rule for the d.c. of products the result for — — ' - — ' 

dx ' dx ' 

dy 
etc. (Thus y — x x x X x, therefore — =, etc.) Prove the rule for 

ax 

positive integers in this way by induction. 

3. The length, width, and height of a cubical block of crystal are given 
respectively by the equations — 

L, = / x (1 + a x r) 
L 2 = h (l + a 2 T ) 

L 3 =/ 3 (1 + a z t) 

where t 19 / 2 , / 3 are the length, width, and height at a temperature at 0° C, 
tfj, a 2) a z are constants, and r the temperature centigrade. 

Find (1) the rate of cubical expansion of the whole crystal per degree 
rise of temperature. (2) The rate of expansion of a block which is I cubic 
inch at 0° C. (3) The rate of expansion of a block which is 1 cubic inch 
at a temperature t. 

Is the rate of (1) cubical expansion, (2) linear expansion, constant at 
all temperatures ? 



CHAPTER VI. 

DIFFERENTIAL COEFFICIENTS OF TRIGONOMETRICAL 
FUNCTIONS. 

§ 36. D.C. of Sin x. 

dfx 11 ) 
We have proved that for all values of n — — = ;/^ (n ~ 1) , and 

we now proceed to deduce expressions and derived curves 
for other functions of x. 

Let y = sin x. 

When sin x, sin 6, and similar trigonometrical expressions 
are used in abstract mathematics, the quantities x, 6, etc., 
invariably refer to an angle of x or radians, and not degrees. 
When degrees are meant, the symbol ° is never omitted. 
Thus sin 2 means the sine of 2 degrees; but sin 2 would 
mean the sine of 2 radians, or 2 x 57*295°. The reason for 
this will appear as we proceed. 

The following practical process will give a tangible con- 
ception of the meaning of the curve y — sin x. 

Draw a circle (Fig. 31) with 1 unit radius, 1 and divide and 
number the circumference, starting at A counter-clockwise, 
into, say, 32 equal parts. Draw a horizontal line OX 
through the centre of the circle. Drop perpendiculars from 
each of the points of division to BOA. Take OX — circum- 
ference of circle = 2 x 3*142 = 6*284, an d divide it into 
32 equal parts. Set up perpendiculars and number them 
1 The unit may conveniently be 3 inches long. 



So 



Graphical Calculus. 



corresponding to the numbers on the circle, through each 
of the points of division, equal in length to the corresponding 
perpendicular to BOA (thus n = n, 22 = 22, etc.; these 
numbers are not shown on Fig. 31), and in the same direction 
as these are drawn. Draw, with great care, a smooth curve 
through all the points thus found. This is the curve jy = sin#, 
for the number of units in the abscissa {e.g. Op) = number 
of radians in the corresponding angle (AOP ), and the ordinate 
pV represents to scale the numerical value of the sine of that 
angle. 




Fig. 31. 

It should be noticed that when we speak of the angle AOR 2 , we refer 
to the whole amount of angle (in this case greater than two right angles) 
included by the arc AP^^ and not the smaller angle included by the other 
part of the circumference. 

Now differentiate this curve graphically. If the work is 
accurately done, a curve will be obtained precisely similar to 
the original curve, but moved to the left by a distance = 1*57, 
which is half the length of one of the loops. 

(Considerable accuracy may be obtained if a large number of points be 
taken, and the scale of the drawing increased.) 



Differential Coefficients of Trigonometrical Functions. 81 

The reason for this peculiarity will now be shown. 

Take two points P^ on the circle (Fig. 31), and find the 
corresponding points PQ on the curve. Draw through ordi- 
nates Yp\ Q^\ We have then — 

Op = arc AP X ; pF = MP l5 etc. 
(V = arc AQ X ; ?Q = NQ X 
therefore mQ = LQ 1 ; Fm = P^ 

therefore — = — — 
rm PiQjL 

But when Q moves up to P in the limit, the figure P1Q1L 
becomes a small right-angled triangle, similar and similarly 
situated to the triangle RxKO, where ORi is perpendicular to 

OPi. Also — - becomes -7, or the tangent of the angle of 
Vm dx 

slope of the curve at the point P. 

since ORx = 1 unit 
Hence we have p y Y x = rR 

Now, pr is evidently = PxRj = 1-57 = — . Hence the 

height of the derived curve at any point P x is the same as the 
height of the original curve at a point 1*57 to the right of 
P ; in other words, the derived curve is exactly similar to the 
original curve moved 1*57 to the left. Its equation must 
therefore be — 

y = sin ( x + * 

Now, if we turn the triangle ORiK round the point O as 
centre, through a right angle into the position OR 2 K 2 , each 
side becomes parallel to a side of the triangle P x OM, and 
since OR 2 = PiO, we have — 

R 2 K 2 = OM = cos x (see § 6) 



82 Graphical Calculus. 

Hence we have — 

^(sin x) 



dx 



= COS X 



For instance, the tangent of the angle of slope of the curve 
y = sin x where x = 1-3 units, suppose, = cos 1*3 radian 
= cos 74-5° = 0*267. 

If we differentiate again, we shall have the original curve 
moved 3*14 to the right; we have, therefore — 

— = sin (x + ir) — — sin x 

ax' 

d(cos x) 
or — - — sin x 

dx 

which result may also be proved independently in the same 
way. This should on no account be omitted by the student. 
Again — 

d\cos x) d'Hsin x) 

— : = =-= = —COS X 

dx J dx 

^ 4 (sin x) 



dx* 



= sin x 



and so on. 



r-m r 1 1 1 ^/(cos x) 

Ihe meaning of the negative sign in the result 



dx 

= — sin x is to be carefully noted. It affords a most instruc- 
tive example of the meaning of derived functions. 

Consider the angle AOPi as being " generated " by the 
line OPi turning round O in the " counter-clockwise " direc- 
tion. As x increases in value, cos x or OM decreases, i.e. the 
increment of cos x (corresponding to a positive increment of 
x) is negative. Thus — 

^(cos x) _ increment of cos x __ negative quantity 
dx increment of x positive quantity 

= negative quantity 



Differential Coefficients of Trigonometrical Functions. 83 

The negative sign indicates that when x increases cos x 
diminishes in the first quadrant. Now, in the second quadrant 
the arithmetical magnitude of cos x increases; but as its 
sign is negative, since M is then on the left of O (see § 2), we 
could no more say that the absolute value of cos x is in- 
creasing under these circumstances, than we could say that 
the value of a man's estate is increasing because his debts are 
increasing. So that, in the second quadrant, ( — sin x) is still 
negative, i.e. cos x diminishes, while x increases. In the third 
quadrant, sin x being negative (since P is below BOA) —sin x, 

or—, is positive, as it should be, because in this quadrant cos x 
dx 

increases algebraically along with x, just as the value of a 

man's estate increases when his debts decrease. In the fourth 

quadrant cos x increases as x increases, because ( — sin x) is 

positive. 

§ 37. Motion of Mechanism of Direct-acting Engine. 

An example of the use of these results is found in the 
investigation of the motion of the mechanism of an ordinary 
steam-engine. Neglect, for the sake of simplicity, the effect 
of the obliquity of the connecting-rod, or assume that the 
crank-pin works in a straight slot perpendicular to the stroke 
of the piston. 

Let OP! (Fig. 32) represent the crank of an engine of 2- 
feet stroke working at 60 revolutions per minute. Required 
the piston velocity when the crank is inclined at 30 , suppose. 

Let a curve be plotted to scale, showing the distance of 
the cross-head from its central position, corresponding to the 
total distance travelled by the crank-pin, starting at far dead 
centre. 

The horizontal scale is to be the same as the vertical. 

Thus at any point P, corresponding to P x on the circle, the 
crank-pin has moved through a distance AP X = Op, and its 
displacement from the central position is clearly OM 2 = pV. 



8 4 



Graphical Calculus. 



When the crank-pin has reached Q x the piston displacement is 
ON = gQ on the other side of centre, and distance moved by 
crank-pin is APjQi = Oa, and so on. Plotting all such values 
on the curve, we clearly get a curve of cosines to a certain 
scale. Now, if we differentiate the curve graphically, the 
meaning of our derived curve will depend on the length we 
take lines such as PL. (The crank-pin is assumed to have a 
constant velocity.) 

L H 




Fig. 33. 



Fig. 34. 



(1) If we take them equal to 1 inch, we shall get a curve 
the length in inches of whose ordinates show the values of the 
ratio — 

small displacement of piston _ piston velocity 

displacement of crank-pin in same time crank-pin velocity 

(2) If we take PL to scale = distance travelled by crank- 
pin in 1 second, the ordinates in inches will show the absolute 
velocity of the piston in feet per second. 



Differential Coefficients of Trigonometrical Functions. 85 

(3) If PL is taken = crank radius, LK will give us the 
velocity of the piston on the same scale as OP^ represents the 
crank-pin velocity. In any case, whatever the length of PL, we 
shall always get a negative curve of sines to some scale or other. 
The height of the derived curve also represents the velocity 
of the piston to some scale, which it is necessary to determine 
from common-sense principles. Thus, suppose the linear scale 
is a quarter full size, and we take PH any arbitrary distance, say 
10 inches. Then HT represents, on the given linear scale, the 
distance that the piston would have moved during the time 
taken by the crank-pin to describe 10 inches X 4 = 40 inches 
if the piston speed had remained the same as it is at the point 
P. Thus HT represents the velocity of the piston at the point 
P to the same scale as PH represents the velocity of the 
crank-pin. From this we can easily find the scale of velocities. 
The crank-pin moves at the rate of 2 x 1 X 3*14 feet per 
second = 6*28 feet per second, and the scale is such that PH 
represents this velocity. The method, therefore, of construct- 
ing the scale is as follows — 

Make P,H, (Fig. 33) = PH, and describe an arc of circle 
with radius PL = 6*28 units, cutting H 5 L in L. 

Mark off the points of the scale as shown, and project 
to P,H,. 

We thus get a scale of piston velocities applicable to the 
derived curve, which would be obtained by making all lines 
such as PH of the given arbitrary length. 1 

Now, it is clear, from what has been said, that distances 
along the line OX may be taken to represent to some scale 
either (1) displacement of crank-pin, or (2) time occupied in 
making that displacement ; for since the velocity of the crank- 
pin is constant = in this case 6*28 feet in 1 second, the same 
distance which represents 6*28 feet on the linear scale along 

1 The ordinates of the curves of velocity and acceleration have been 
reduced in the figure owing to want of space. The student should draw 
a larger figure for himself. 



86 Graphical Calculus 

OX will also represent i second. Thus, if we differentiate the 
derived curve, bearing this in mind, 1 we shall be able easily to 
find the scale of accelerations applicable to the curve thus 
produced (§ 16). 

Thus, suppose P^IT represents 0*25 second. It is clear that 
H v r = /"P" represents a rate of change of velocity of IVV 
(measured by the velocity scale already constructed) in 0*25 
second, i.e. four times that change of velocity in 1 second. 
Thus, suppose H r T\ when measured by the velocity scale, to 
represent 4-52 feet-per-second. Then the acceleration scale is 
such that / n P n represents a rate of change of velocity of 4*52 
feet-per-second per i of a second, i.e. 18*08 feet-per-second 
per second, and we can proceed as before to construct a 
complete scale with which to measure accelerations on the 
second derived curve. 

The student who desires to understand the subject thoroughly should 
on no account omit to perform the complete process himself, and think 
out himself ab initio all the principles involved in the construction of his 
own scales. It need not be pointed out to him that the whole process is 
utterly useless unless he can construct exact scales for himself by which to 
measure his curves. He should be able to alter his scale at pleasure, in 
case he has not room enough to adhere to one. 

The derivation of the second derived curve is of the 
highest importance in calculations respecting the inertia of 
moving parts in high-speed engines. It will be found, in the 
process of graphical differentiation, unless very great care be 
taken in the exact determination of a large number of points 
on the curves, that great and cumulative errors may be made 
in the drawing of the tangents to the curves. For this reason, 
other and more accurate methods are preferable where it is 
possible to find them. In particular, very simple and accurate 
methods are known for determining the curves here found by 
the process just explained. Those processes also take account 
of the varying obliquity of the connecting-rod. We might 

1 It is best to mark off O v X x in seconds or half or quarter seconds. 



Differential Coefficients of Trigonometrical Functions. 87 

have done the same by a modification of the construction for 
determining the curve of displacements ; but if we had done so, 
the process would not have corresponded with the algebraical 
investigation to be given shortly. A proof is here given which 
shows the real though obscure connection of the following 
process with that of graphical differentiation. The student 
should not fail to perform the operation by both processes 
and compare them. 

Divide the circle representing (to scale) the path of the 
crank-pin into a number (32) of equal parts. Draw the line 
of centres, and put in the centre lines of the various positions 
of the connecting-rod with one end on the line of centres and 
the other end on the circle at the points of division. Produce 




Fig. 35. 



the connecting-rod, if necessary, till it cuts the vertical 
through O in H; then OH represents the velocity of the 
piston to the same scale as OP represents the velocity of the 
crank-pin. For consider two infinitely near positions of 
the crank-pin P and P^ Draw in the two positions of the 
connecting-rod QP and Q^. Draw a horizontal P 2 N, and 
with centre Q and radius QP describe a small arc of a circle 
PN. Then NQQ^ is a parallelogram, for Q^ = QN, and 
NP X is parallel to QQ X . Therefore NP X = QQ r Thus, while 
the crank-pin describes PP 1? the piston describes NP X . Now, 
in the limit when PP X is indefinitely small, the small figure 



88 Graphical Caladus. 

NPPj becomes a triangle of which PP X is perpendicular to 
OP, PN to PH, and NP X to OH. Hence if we turn the small 
triangle NPPx through a right angle round the point P in the 
right-handed direction, each of its sides will be parallel to one 
of the sides of the triangle POH. Hence we have by similar 
triangles — 

PiN = OH 

PjP " OP 
. velocity of piston at point P _ OH 
velocity of crank-pin OP 

i.e. on the same scale as that on which OP represents the 

velocity of the crank-pin, OH represents that of the piston. 

Hence, plotting the values of OH on a base representing the 

path of the crank-pin unrolled, we get a curve of velocities 

of the piston which is in practice more accurate than that 

obtained by direct differentiation. 

In the same way it may be proved that if HM be drawn 

horizontal to meet OP produced, and ML vertical to cut QP 

in L, and LK perpendicular to the connecting-rod, then OK 

represents the acceleration of the piston to the same scale as 

OP represents the radial acceleration of the crank-pin, viz. 

z> 2 
wV, or — , where o> represents the angular velocity of the crank 

in radians per second, and v the linear velocity of the pin. A 
curve plotted on a similar base to the preceding and vertically 
underneath it, shows the value of the piston acceleration. 

The student should not fail to draw this curve, and to demonstrate to 
himself that it is the same curve as would have been derived by graphical 
differentiation from the curve of velocities, as obtained by the method of 

Fig- 34- 

It is impossible for any one to properly appreciate the extremely 
instructive points involved in these constructions without thoughtfully 
drawing the curves to scale. 

Now, the algebraical investigation of the same thing is the 
exact counterpart of the process first described, neglecting 
the effect of obliquity of the rod. 



Differential Coefficients of Trigonometrical Functions. 89 

Let y represent the displacement of the piston from its 
central position • 

x the angular displacement of the crank, starting from A 
(Fig. 32) in radians; 

And r the radius of the crank-pin circle. 



Then y = 


f cos X 


dx 


— r sin x 


dy 

rdx 


— sin x 



Now, rdx = distance moved by crank-pin, 1 while the crank 
describes the angle dx radians (for, as in § 6, since angle 

= — - — , therefore arc = angle X radius) : dy = distance 
radius 

moved by piston in same time. 

dy 
Hence — - = ratio of velocities of piston and crank-pin. 

rdx 

Take any particular value for the angle, say 30 = -? 
= AOP. 

TV 

Velocity ratio at P = — sin -? = — J 

i.e. the piston is moving backwards half as fast as the crank- 
pin is moving. 

§ $%. D.C of y = Sin- 1 x. 
From the result already obtained — 
d(sin x) 

- A - : ' = COS X 

dx 

combined with the principle deduced in § 23, we can at once 
find an expression giving the height of the first derived curve 
of the curve y = sin" 1 x. 

1 See note (7), p. 194. 



9^ Graphical Calculus. 

As already shown, this means " y is equal to the angle 
(in radians) whose sine is x" 

Let the curve OQiPj in Fig. 36 be Y = sin X. Rotate 
this curve through a right angle into position dotted, and 
reflect it, and we obtain the curve x - sin y, or, as it may be 
written, y — sin -1 x. 

x is thus geometrically substituted for Y, and y for X. x and X are 
both plotted horizontally. 

This is the curve OPQP 2 . Consider the point P. 

-7- = limit of =777 when Q is infinitely near to P, 
dx PN ^ J ' 

= limit of 



N X Q L 

_dX _ 1 1 

dY cos X cos j 

for every value of X = corresponding value of y\ Tins result 
is perfectly satisfactory, and is all we require if y is to be the 
independent variable ; but if x is, as usual, the independent 
variable, we can immediately find the value of this in terms 
of x by substituting ; thus — 



= ± 



cosj> Vi -sin 2 jy Vi 

For the derived curve, therefore, we get— 



VI —x j 



which is usually written without the double sign, because 
writing a double sign may be considered as part of the process 
of finding a square root. 

The shape of this curve is shown in the figure. It consists 
of two infinite branches as shown. At P the slope is p y F y ; but 
at P 2 , of which the abscissa is the same as that of P, the slope 



Differential Coefficients of Trigonometrical Functions, gi 




FlG« 36. 1 

1 See note (8), p. 194. 



92 Graphical Calculus. 

is />T 2 \ The meaning of the double sign is thus rendered 

evident. Each of these two branches approach the lines SA, 

but it is clear they never actually touch it in finite space \ for 

dy 
at the point S, ~- is infinitely great, and though by taking 

the point P near to S we can make the distance of the point 
P r from the line SA as small as we please, yet the ordinate 
of P^ becomes enormously great, and the actual co-ordinates 
of the point S x would be (i, oo.) When a line and a curve 
have this relation to one another, i.e. the curve continually 
approaches as near as we please to the line, but never 
actually meets it in finite space, the line is said to be an 
" asymptote " * to the curve. These asymptotes are of great 
importance in the general tracing of curves. In general, both 
co-ordinates of the point of contact are infinite. 

D.C. of Cos -1 ^. — It is clear that by moving the vertical 

curve downwards through a distance = -, so that the point 

2 

S is on the line OX, we shall obtain the curve y - cos -1 ^, 
since we obtain the curve Y = cos X by moving the horizontal 
curve to the left through the same distance. Now, it is clear 
that this does not in the least alter the derived curve, so that — 

^(cos -1 x) _ d(§m~ l x) i 

dx dx ~ Vi — x 2 

as may be proved independently, thus — 

y = COS" 1 X 

x = cosy 

dx . i 

— = — sin y = + vi —x l 
dy 



dx ~~ a/i —x 1 
the same expression as before. 

1 From three Greek words, signifying "not falling together." 



Differential Coefficients of Trigonometrical Functions. 93 

It will, of course, be seen that neither of the curves y = 
sin" 1 ^nor y = cos" 1 x can have an abscissa > 1 or < — 1 ; 
for there is no possible angle which has a sine or cosine 
> 1 or < — 1. The same thing may be seen in the equation 
to the derived curve; for if x becomes greater than 1, say 2, 
we have — 

1 

* -3 

an imaginary expression, for it is impossible that a negative 
quantity should have a real square root, since the square of 
any real quantity, positive or negative, has a positive sign. 



§ 39- 
In a similar way the d.c.'s of tan x, cot x, sec x, and cosec 
x can be obtained. The principles involved have in previous 



Q/ kcvtx P, 




Fig. 37. 

sections been fully explained, and as these can also be easily 
obtained from the dc.'s of sin x and cos x, we shall merely 
give brief geometrical proofs. The student should in all cases 
draw the actual curves. 

Make Ox\ = 1 inch (Fig. 37). Draw tangents at A and D. 
Consider the point B on the circle. 

Let AB, or the angle AOB in radians, = x. 

Then AP = tan x, PQ = A (tan x), BC = Ax 



94 Graphical Calculus. 

d(iznx) .. . PQ 
A__ = hmit of — 

PQ PN 

" pn'bc 

PQ OP 
PN ' OB 
PQ OP 

pn'oa 

= sec 2 x in limit 
since angle QPN = angle AOP 

Also DPj = cot x, PjQ! = A cot x. 

d(coi x) ,. . , P^ 

-*r = hmit of # 

BC 

= limit of -^.^ 

MQ X BC 

Q1P1 OQi 

MQ 2 ' OD 
= — cosec 2 x 

Again, sec x — OP, A(sec #) = NQ. 

-^- — - - limit of -^ 
*fo BC 

NQ PN 

pn'bc 

NQ OP 
PN'OA 

= tan ^ . sec # 

Again, cosec # = OP D A(cosec x) = — MP X . 

d(co*ec #) _. . r MP, 

- v — = limit of - — - 1 

dx BC 



Differential Coefficients of Trigonometrical Functions. 95 

MP MQi 
" MQ X ' BC 

mq/od" 

- —cot x . cosec x 

Exercises. — (These exercises are of the highest importance. ) 
Prove each of these results from the d. c.'s of sin x and cos x 
on the principles explained in Chapter V. in the following 
manner : — 

//(tan x) d f sin x \ cos 2 x + sin 2 x 
dx dx\ cos x / cos 2 x 

1 2 

= — = sec x 



COS" X 

Prove also the following results by the same method as 
that explained for y - sin -1 .*, drawing the curve in each 
case. 

d(tan~ l x) _ 1 
dx 1 + a? 

d(cot~ 1 x) _ 1 



dx 1 -f- x 2 

d(sec~~ 1 x) _ 1 

dx xs 1 x -1 

//(cosec - 1 x) _ 1 

dx x vx 2 — 1 

Thus, ify = tan" 1 x } then x = tan y. 

dx 

-— = sec 2 y — 1 + tan 2 y = 1 + # 3 

#7 

^ _ I 

dfc 1 + ^ 2 

Prove these results graphically by tracing the curves and 
inverting them. 



CHAPTER VII. 

DIFFERENTIAL COEFFICIENTS OF LOGARITHMIC FUNCTIONS. 

§ 40. D.C. of Log x. 

We will now consider the curve y = log x. A remark 
similar to the one we made in defining the meaning of such 
expressions as sin x applies here, viz. that in abstract mathe- 
matics log x with no suffix signifies, not the ordinary logarithm 
as found in log tables, but the " natural " logarithm to base 
" e " where e is the value of the infinite series — - 

1 +!+ 1+ 1+ . . . = 271828 . . ., etc. 

which is the value which the expression f 1 + - J assumes 

when n is infinitely great. The student cannot hope to 
understand this fully unless he be acquainted with the 
algebraical theory of logarithms, which is found in any fairly 
advanced book on algebra, such as Hall and Knights' " Higher 
Algebra," He may, nevertheless, obtain approximate values 
of the natural logarithm of a number by multiplying its 
ordinary logarithm (to base 10) by the log of 10 to base e, 
viz. 2*303 about. 

Calculate in this way the natural logarithm of 0*25, 0*5, 075, 
1-25, 1*50, 2'o, 3'o, 4*0, 5'o, 7-5, 10. Plot points whose 
abscissas are the numbers here given in inches or other units, and 
ordinates the calculated logarithms. Carefully draw a smooth 
curve through these points. This curve crosses the line OX at 



Differential Coefficients of Logarithmic Functions. 97 

a point whose abscissa is 1 ; for with any base whatever 
log 1 = 0. 

On the left of point (1,0) care must be taken : thus from 

the tables we can find log 10 0-5 = 1*69897, which for our 



s 



-H* 



/ 



/ 



y 




Fig. 38. 



98 Graphical Calculus. 

purpose is practically equivalent to 0700, since we cannot 
plot correct to y^- inch. 
Hence we have — 

logio 0-5 = - 1 + 070 = -0-3 
hence hyp. log 0*5 = -0*3 X 2-303 = —0-69 

and similarly for the other points. 

Draw a curve through the points. This curve is shown in 
a full line in Fig. 38. 

Now differentiate this curve graphically. The general 
shape of the curve obtained will be as shown in the lower 
part of Fig. 38. Take a number of points such as P' on 
the curve, measure with a decimal scale p y P y and O y p y , multiply 
their lengths together, and the result will be found to 
be always 1 if the work is accurately done. Its equation 

must therefore be xy y = 1, ory = -. 

x 

Another way of exhibiting this fact very clearly is to take 
a number of points P on the primary, through which erect 
pT perpendicular = 1 inch. Join OT. Then OT will be found 
parallel to the tangent PR at P. 

Exercise. — The whole curve may therefore be drawn by 
the method explained in § 14. Draw it in this way, and 
compare it with the curve just plotted. 

These exceedingly important facts may be proved alge- 
braically as follows. Consider another ordinate $Q near pT, 
distance h from it. 

Let Op = x ; p? = log x. 

Then Oq = x + h 

?Q = log (x + h) 

Therefore, with the usual notation— 

Ay __ log (x + h) —log x 
Ax h 



Differential Coefficients of Logarithmic Functions. 99 
which, from the nature of logarithms 

x n \ x J 

Write n instead of f . This becomes— 
h 

-logfi+i) 

When ^ or A* diminishes indefinitely, it is clear that 
n increases indefinitely. When, therefore, this takes place, 

log ( 1 + - J becomes, from the definition above, log e = 1. 

At the same time, when A* or h dwindles indefinitely, — be- 

Ax 

before, 



comes -r, or height of derived curve ; hence -7- = y = -, as 
dx ax ^ x 



§ 41. Illustrations. 

Some interesting and instructive results may be derived 
from these equations. On the same base as before, plot the 
logarithms as found in ordinary tables. A curve will be 
obtained similar to the other in general character, but flatter. 

As we have seen, it is — as high at all points. It is the lower 

dotted curve in the figure. Its equation is y = fx log e x. 1 
It crosses the line OX at the same point as the other. Its 
slope at this point may be approximately found from the 
tables, for we have — 

log 10 1 = 0*000,0000 

log w roooi = 0-000,0434 

L. cf G. . c L , . 

1 See note (9), p. 194. 



ioo Graphical Calculus. 

Hence at this point Ay = 0*000,0434 

Ax = o'oooi 

Ay 

— -, which, when Ax is so small as 0*0001, will be very 

nearly = — = 0-434. 

This is also evident from the equation — 
My = fi log, x 

d y * t it o\ 

— = - (see § 28) 
ax x 

which (when x = 1) = /jl. 

It is interesting to notice that, in an ordinary book of logarithms, the 

height of the derived curve of the curve of ordinary logarithms is given 

by the side of the tables, so as to enable any one using the tables to 

"interpolate." This height is called "difference" in the tables. The 

principle made use of in the calculation of intermediate logarithms is 

dy dy 

8y = — • dx. The value of — is given as a "difference." 
ax dx 

We can obtain another curve of the same character by 
plotting the lengths taken from a slide rule on the same 
base. This is the upper dotted curve in Fig. 38. The 
graduations of a slide rule are ruled proportional to the 
logarithms of the numbers engraved on the rule, so that 
addition and subtraction on the rule, which are easily per- 
formed mechanically by sliding one scale over the other, are 
equivalent to multiplication and division respectively. On 
the ordinary small " Gravet " rule, log 10 is represented by 
12*5 cm. = 4*921 inches. 

. 4*921 

Hence this latter curve is = 2*135 times as high 

2*302 

as the e curve, and its slope at the point (1,0) is 2*135 (§ 2 ^)« 

The result we have obtained may also be written 

dx 

— = log x -)- c. In this form it is extremely useful to the 



Differential Coefficients of Logarithmic Functions. 101 

engineer in enabling him to find the work*done by a gas 
(such as air) in expanding isothermally or at constant tem- 
peratures. This will be fully considered in the next chapters. 



§ 42. D.C. of e x . 

By inverting the curve of logarithms, as explained in § 23, 
we can prove a result of great importance. 

The curve P x (Fig. 39) is the curve Y = log X. Rotate 
it about point O into the position dotted, and reflect on OY, 
and we get a curve whose equation might be written con- 
formably with those of sin x and cos x 9 etc., y = log" -1 x, or 
y is the number whose logarithm is x. 

It is usual, however, to write the equation y = e*, for e x is 

obviously, from the definition, the number whose logarithm 

to base e is x. 

If the student cannot understand this, he is referred to any book on 
algebra which contains a chapter on logarithms. 

If this curve is differentiated graphically, the result will 

be a curve which is exactly in every respect like the primary 

curve. In other words, the peculiarity of this curve is that 

if the tangent at a point P be produced so as to meet OX 

in S, then, wherever P is on the curve, Sp will be exactly 

1 inch, for the triangle S/P is evidently exactly equal and 

similarly situated to the triangle we should have drawn for the 

point P in differentiating the curve in the ordinary way. This 

is expressed by saying that the " subtangent " is constant, Sp 

being the subtangent. 

This result may be proved as follows : — 

If y = e x 

then x == log y 

dx 1 

'■ dy ~~ y 



102 



Graphical Calculus. 



An interesting property of this curve is that, if a series 
of abscissae are taken in arithmetical progression, the corre- 
sponding ordinates are in geometrical progression. The 




Differential Coefficients of Logarithmic Functions. 103 

student should prove this algebraically from the equation to 
the curve. 

The whole curve should be obtained by this method, 
by taking ordinates 0*25 inch apart. The ratio of the 
ordinates will be ^°" 25 , the value of which must be cal- 
culated, and the successive ordinates found geometrically by 
a construction similar to that of Fig. 3. All logarithms can 
be graphically obtained from this curve by measuring the 
abscissae corresponding to an ordinate whose length = number 
whose log is required. 

The whole of the results in this chapter and the last must 
)e thoroughly learnt off by heart. The student who wishes 
o proceed with the subject will save himself much time and 
anoyance by making himself perfectly familiar with them at 
he outset. It is not too much to say that one-half of the 
iifficulty usually met by elementary students of the integral 
:alculus is due to an imperfect knowledge of these few simple 
:esults. The student can best learn them by deducing them 
for himself once every day, and constantly picturing to him- 
self the curves representing the functions and their differential 
coefficients. He thus obtains a practical and real familiarity 
with the functions, such as he could not get by studying the 
symbols only. Unless he is gifted with an exceptional 
memory, he will find even the few here collected difficult to 
remember otherwise than by understanding what they mean. 
The results should be as familiar forwards as they are back- 

/ dx 
wards ; e.g. he should know that I — = sec -1 x just as 



xv x l — : 



dtfsec 1 x) 
well as that — - 



dx xs!x l — 



1 



i04 



Graphical Calculus. 



Direct. 


Inverse. 


Function. 


Diff. coefficient. 


Function. 


Diff. coefficient 


sin x 


fix 31 - 1 
cos *■ 

— sin x 
sec 2 .r 

— cosec 2 x 
sec or tan .r 

— cosec x cot .r 
i 

X 


sin -1 ;r 

COS" 1 X 

tan -1 .r 
cot -1 .r 
sec -1 x 
cosec -1 ^- 


4- ' 


COS X 


VI -*- 2 

+ ' 


tan x 
cot X 


\l I — X 2 
I 


I +X 2 

— I 


1 + x 2 

I 


sec X 




cosec .r 


.rvj 2 — i 
— i 


log .r 


x*Jx 2 — I 



Examples. 

i. What is the equation to the inverse curve of the lower dotted curve 
in Fig, 38 ? Is the subtangent constant in this curve ? Is the first derived 
curve like the primary curve ? Prove your answer graphically and 
analytically. 

2. Assuming the result for the d.c. of **, *prove algebraically by 
inversion the result for y = log x. 

3. Differentiate y = a x . 

See note (10), p. 195. 



CHAPTER VIII. 

DIFFERENTIATION OF A FUNCTION OF A FUNCTION OF A 
VARIABLE WITH RESPECT TO THAT VARIABLE. 

§ 43- 
We have considered, in the preceding chapters, the process of 
differentiation of simple functions of a variable x (such as 
sin x, log x, etc.) with respect to that variable — that is, the 
relative magnitude of the change produced in the value of the 
function by a small change in the value of the variable. 
Now, this small change in the value of the variable may 
have been itself produced by a change in some other variable 
(z, suppose), on the value of which x depends, and it is often 
necessary to know the ratio between a change in the value of 
the given function of x and a small change in the value of z 
(which latter produces a certain change in x, and in conse- 
quence a change in f(x), the function to be differentiated). 
In other words, we have to differentiate some function of x 
(say log x) with respect, not to x, but to z, i.e. to find the 

d(\og x) 
value of . Of course, this would not be possible 

dz 

unless there were some relation subsisting between x and z, 

such that x takes up a definite value corresponding to any 

given value of z (see the note at the end of § 23, on p. 49). 

As the meaning of this process is usually very confusing to 

the beginner, and as it is important that he gets clear ideas on 

it, we shall illustrate it by an everyday example. 



106 Graphical Calculus. 

Suppose a tradesman starts in business for himself at the 
beginning of the year 1870. At the beginning of that year he 
earns profit at the rate of ^200 per annum, or about in. per 
day, or is. ^\d. an hour. 

Suppose this rate of profit gradually and regularly increases 
by ^20 per annum every year, so that, for instance, in the 
middle of 1870 he is earning \\s % 6d. per day, or ^210 per 
year; and at the beginning of 187 1 he is earning ^220 per 
year, or about 12s. a day. It is clear that his average rate 
of profit throughout 1870 has been ^210 per annum, which 
sum also represents his total earnings for the year. At the 
beginning of 1872 he is earning ^240 a year, and so on (i.). 

Let the current rate of profit at any time be denoted 
by £z per annum, and suppose his current rate ' of living 
expenditure at the same time is given by £z$ per annum, 
denoted by y (ii.). 

It is required to find the rate per annum at which his 
rate of living expenditure is increasing. 

This example can most easily be understood by following the curves on 
Fig. 40. The "dimensions " of this rate of increase will be "pounds-per- 
annum every year," in the same way as the dimensions of an acceleration 
are " feet-per-second every second." It would be incorrect to measure 
this rate of increase in " pounds-per-annum," because " pounds-per- 
annum " are the dimensions of an income or annual expenditure, and not 
a rate of annual increase of income or of annual expenditure (cf. Fig. 17). 

Let y be his rate of living per year, and z his rate of earning 
profit (both in pounds per annum), at a time represented by x 
years counted from the beginning of 1870. 

It is evident that the relation between y and z is — 

y = z* . . . . (a) 
This being the algebraical expression of supposition (ii.) 
above. 

Also we have — 

Z = ^200 +;£2C* , . .(b) 

which expresses supposition (i.). 



D.C. of a Function of a Function. 107 

dy 
We have then to find the value of — . Now, from equation 

dx 

(a) we can (§ 34) easily determine 7-, or his rate of increase 

dz 

of expenditure per £,\ increase of income; 1 but this is not 

dz 
what we want. Also from (b) we can find — , or his rate of 

increase of income per year (§ 16); but neither is this what 
we require. 

Now, from these two equations, (a) and (£), we can obtain 
another involving only y and x, for we can substitute ^200 
+ J^2ox instead of z in the equation — 

y = 2* 
This process is called " eliminating z between (a) and (&)" 
We thus obtain — 

y = (;£200 + ^20.*)* . . ..(c) 

Here we are fixed, for we have hitherto proved no rule which 
will enable us to differentiate this expression with respect to 
x. We have, in fact, come to a point where we must 
differentiate a function (viz. the power J) of a function (viz. 
200 + 20.x) of a variable (x) with respect to that variable. 

If the student has followed the previous reasoning carefully, he will 
probably suspect that we shall find what we require by multiplying together 
the two d.c's already found ; that is — 

dy dy dz 

but he must be very careful to notice that he has no right whatever to 

take this result as proved merely because ^ and -r look like fractions. 

dz dx 

He should know already that dx, dy, and dz are not quantities to which 

dy 
1 If we take ;£ioo as the unit, — represents the amount by which his 

dz 

rate of expenditure increases per ^100 increase of income. In this case, 

however, z would represent the profit in hundreds of pounds per annum, 

and we should take account of this algebraically by modifying equations 

(a) and (b) according to the units we are working in. 



io8 



Graphical Calculus. 



definite values can be assigned, and therefore to cancel out one dz with 
another without inquiring into the meaning of the process is an operation 

dy 

which is quite as illegitimate as it would be to cancel out the d in , and 

to put -T- = — . In certain cases the latter might be true, .but in the great 
r dx x 

majority of cases it would not be. It would signify that the tangent at a 

point P of a curve passes th ough the origin O, which is obviously generally 

untrue. 

The student's aim should be to grasp the meaning underlying all these 

symbols. He should never perform algebraical operations of this kind in 




a haphazard fashion without making himself acquainted with the principle 

dy dy dz 

involved. In this case it is perfectly true that — = — X — , but it requires 

dx dz dx 

proof before it can be accepted, and it is only to be taken as another 

analogy between the laws relating to differential coefficients and those 

relating to fractions. 

Draw the two curves representing relations (a) and (b) 
as shown in Fig. 40. (The curves in the figure are not 
drawn to scale.) Notice that curve {a) does not involve the 
idea of time, but simply shows the expenditure corresponding 



D.C. of a Function of a Function, 109 

to any income. Also that, since, during the period under 
consideration, the rate of profit is always greater than ^200 
per annum, we have nothing to do with the dotted part of 
the curve. 

Curve (b) shows, in the length of its ordinate, the income 
corresponding to a time given by the abscissa. To obtain a 
curve showing time — annual expenditure — we must combine 
the abscissae of (a) with the corresponding ordinates of (a). 
Thus, consider a time two years after January, 1870, i.e. 
January, 1872. The income is given by /iP x . Transfer this 
to O2A as shown. Then p 2 Y 2 gives the living expenditure at 
this date. Take a base, 3 X, divided exactly like O x X, and 
transfer the ordinates such as/ 2 P 2 to/ 3 P 3 , where 3 / 3 = 1 p 1 . 
This curve, when drawn, is the result of graphically eliminating 
z between (a) and (b). 

Consider corresponding ordinates, Q 1? Q 2 , Q 3 , adjacent to 
P 1? P 2 , P 3 , where Q 2 , Q 3 are obtained from Q 1} exactly as P 2 , P 3 
were obtained from P x . 

Then clearly — 

LQj = P 2 M 
MQ 2 = NQ, 
P 3 N = P X L 

Hence we have — 

NQ 3= MQ 2 = MQ 2 LQ, 
P 3 N P X L P 2 M * PjL 

From the way in which the curves were constructed, this 
is true wherever Q may be. Now, when Q x approaches P 1? so 
that P X L dwindles indefinitely, it is clear that all the other 
quantities in the above equation do the same ; and when this 
is the case, the equation becomes — 

dy dy dz 

dx dz dx 

for the three ratios which are contained in the equation 



no Graphical Calculus. 

become respectively — , —■, and — , whatever actual values 
dx dz dx 

the quantities denoted by NQ 3 , P 3 N, P 2 M may have, pro- 
vided always these are infinitely small (since the part of the 
curve along which Q may move consistently with this con- 
dition is an infinitely short straight line, as already explained 
in § 13 ; see also note on p. 48). 

Differentiating all three curves, then we see that any 
ordinate of (V) X corresponding ordinate of (a!) = correspond- 
ing ordinate of (<r f ). The criterion of correspondence is, of 
course^ not the same as that in the case of the curves multi- 
plied together in the ordinary sense. Thus pi Pi, / 2 'P 2 ', P%P%, 
are corresponding crdinates, although 0(pl is not = 2 ' / 2 \ 

It is clear that the curve (c') in this case is represented 
by- 



== |-(200 + 20X) ~s X 20 = 



\/200 + 20X 



8 44. 

(i.) On the same principle, we can differentiate such ex- 
pressions as (sin x) 2 . 

The curves in this case are — 







y = z 2 . 
z - sin x 


■ ■ 00 


By eliminating z, 


we 


obtain — 




Here — 




y = (sin x) 2 . 


• • (0 






dy 
dz 


• • 00 






dz 

-7 = cos X. 

dx 


. • w 



D. C. of a Function of a Function. 1 1 i 

Hence — 

dy dy dz 

-~ = -• • — - 2Z COS X 

dx dz dx 

= 2 sin x cos x . . t (V) 

The letters denoting the equation correspond to the same 

letters in the illustration. 

(ii.) A frequent application of the same principle occurs 

x 
in the differentiation of such expressions as sin -. Here we 

a 

x 
might be tempted to think that the d.c. was cos -. But it 

must be carefully noticed that this would be the d.c. with 

x 
respect to -, and not to x (see Examples II. at end of Chapter 

in.). 

x 
Here z = - • . . . . (/>) 
a 

y - sin z . . . • , (a) 

dy dy dz i x 
— = — . — = - cos- 
dx dz dx a a . 

x 
The mistake in this case arises from the fact that - is, for 

a 

convenience, not usually enclosed in a bracket, although it 

might be if desired. 

(Hi.) Take another case : y = log (sin x). 

Let z = sin x . . . . (b^) 
thenjy = log z . . . (a) 

dy _dy dz 
dx dz ' dx 

i i 

= - • cos x — — cos X 

z sin x 

= cot X 
(iv.) Take a more complicated case : y - (e* cos x~) n . 



112 Graphical Calculus. 



Let z = e* cos x . (a) 

y = z n . . . . {b) 

dy __dy dz 
dx dz dx 
dy 



dz 



= nz 



dz 

-j- must be found by the rule for products of functions 

given in § 30, thus — 

dz d(cos x) , die?) 

— = e c ^— i + cos x -*-*■ 

dx dx dx 

= — e° sin x -f e x cos x 

= £ x (cos x — sin x) 



Hence — 

dy 
dx 



n(e x cos x) n ~ l X e*(cos x — sin x) 



After a certain amount of practice, the student will find 
that he is able to dispense with the z substitution, and to 
write down the result without any intermediate step. 

(v.) A difficulty arises to beginners when they have to 
differentiate such an expression as, say, q B with respect to x. 
They are tempted to write down as the result $q 2 , forgetting 
that this is the d.c. with respect to q^ and not to x. They 
are often unable to trace the meaning of differentiating q 3 , 
which does not appear to contain x, with respect to x. If so, 
they should read again the note at the end of § 23, and 
remember that there could not be such a thing as a d.c. of 
<f with respect to x unless a relation such as is there de- 
scribed subsisted between q and x. 

-y^ is therefore 3^ . — 
dx dx 

dy _ dy dq 

' ' dx dq ' dx 



D.C. of a Function of a Function. 1 1 3 

The rule, therefore, is as follows : — 

To differentiate any function of a quantity enclosed in a 
bracket with respect to a variable x {e.g. cos (log x)) — 

(i.) Differentiate the expression, treating the whole quantity 
in the bracket as an independent variable. This would give 
us —sin (log x). 

(ii.) Multiply this by the differential coefficient with respect 
to the variable of the quantity enclosed in the bracket. Here 

the d.c. of (log x) is -. Hence — 

d{cos (log x)} __ — sin (log x) 

dx x 

If there are two or more brackets enclosed one within the 
other, it is easy to see by induction that we must first treat 
the whole of the outside bracket as an independent variable, 
and proceed inwards, treating each bracket in turn as the 
independent variable, multiplying all the successive results 
together. Careful attention to the following example will 
enable the student to understand this. 

Letj> = [log {log(sin **)}]*. 

(i) Differentiate as though the quantity contained in the 
] brackets were an independent variable. This gives us — 

/* [log {log (sin^)}]"- 1 

(2) From our rule, it is clear that this must be multiplied 
by the d.c. of the quantity contained in the [ | brackets. 
Hence we have, as it were, to start the same process over 
again, absolutely neglecting everything outside the [ ] brackets. 
This, according to our rule, will involve treating the quantity 
in the { } brackets as an independent variable. Thus far we 
have — 

«[log {log (sin**)}] 71 " 1 X 



{ log (sin e x )\ 

1 



U4 Graphical Calculus. 

Now, in order to obtain the d.c. mentioned in (2) above, 

we must multiply : — by the d.c. of the quantity in the 

r ' {log (sin**)} 

{ } brackets, which in turn involves treating the expression 
in the ( ) brackets as independent variable. This gives — 

n\\og {log (sin ^HT 1 " 1 X 7: ~ rr X tt^— 

L 5 l 5V /yj {log (sin <?)} (sm<f) 

which we must then multiply by the d.c. of the quantity in 
the ( ) brackets. This involves treating e* as an independent 
variable. (The student is apt to stumble at the last step, 
because e x is not enclosed in visible brackets.) Finally, the 
whole expression must be multiplied by the d.c. of e* with 
respect to x. The whole expression is then 

//[log {log (sin ^)}>~ 1 X y - — X -7^— Xcos f X ? 

)J {log (sin f)} (sm<f) 

The student should not be satisfied till he can write out 
any complicated result like this at sight, without any sub- 
stitutions. He must learn to fix his attention on each bracket 
in turn, treating it quite apart from anything else, and regard- 
ing the next bracket proceeding inwards as the independent 
variable. If he finds it impossible at first to avoid getting 
the thread of his thoughts entangled among the brackets, he 
should get a separate piece of paper and cross each bracket 
out as k is done with. He will thus find an apparently 
extremely complicated expression quite simple to differentiate. 



§ 45. Applications. 

The application of this rule is the source of much of the 
difficulty which the student meets with in applying elementary 
calculus to science. Differential coefficients of quantities are 
sometimes treated of with respect to variables, with which the 



D. C. of a Function of a Function. 115 

quantities have no apparent connection. New variables are 
often arbitrarily introduced, and d.c's assumed with respect 
to them; so the student is quite bewildered by the multiplicity 
of symbols. He is again reminded that the very existence of 
a d.c. of any quantity with respect to a variable involves the 
existence of a definite relation such that, other variables being 
constant, the assumption of a particular value by one fixes 
the value of the other. 

For instance, suppose that each of the following variables, 
(#), (3), (c), (d), (e), etc., are exclusively dependent on (A), the 
temperature during the winter : — 

(a) The number of unemployed workmen. 

(b) The demand for overcoats. 

(c) The amount of railway traffic. 

(d) The sale of skates. 
(e) The death rate, etc. 

We are assuming that we have curves given, representing 
the value of each of these variables, corresponding to values 
of (A). Derived curves could be obtained representing their 
rates of increase or decrease per degree-rise of the thermo- 
meter. From these curves we could find a relation such as 

d(b) c dip) d(b) d(A) .. _ . . 

-V-r, for -77- = — -7 . -7— , although the demand for overcoats 

d(e) d(e) d(A) d{e) 

might have no apparent connection with the death rate. 

Or, again, we might introduce the arbitrary variable time, 

although in our original curves the idea of time did not 

enter ; but, in order to make such a relation as — have any 

determinable value, we must have given a curve showing the 
relation between any one of these variables and the time. 
Suppose the primary and first derived of the time-temperature 
curve had been given. Then we have, say — 

d(d) = d(d) dA 

dt dA ' dt 
and so on. 



Ii6 Graphical Calculus. 

Direct-acting Engine. — We have already had a disguised 

example of the application of this principle in the case of 

the engine in § 37, which we now proceed to explain more 

fully. What we actually wish to find in the problem is the 

velocity of the piston, and this, as we have seen in § 16, is 

the first derived function of the time-displacement relation, 

dp 

or -f . 

dt 

Now, the geometrical relation between crank angle 6 and 

piston position p furnishes us with the means of finding the 

dp 
value of -j for any value of 0. This quantity (neglecting 

obliquity) we have seen to be — r sin 6. Hence we have 
only to multiply by the corresponding value of the relation 

— (i.e. the height of the first derived of the time-angle curve) 
dt 

A * £ A d P f d P ^ d6 

m order to find-, for - = ^ -. 

Now, we know from the data of the problem that the time- 
angle curve is a sloping straight line, since the motion of the 
crank is a uniform rotation, i.e. the amount of angle described 
is proportional to the time ; hence the first derived is a hori- 
zontal line, or the " angular velocity is constant." The height 
of this first derived is given in the problem, for we are told 
the crank turns at 60 revolutions a minute, or 27r radians per 
second. Hence for the time-piston displacement first derived 
curve we have — 

dp dp a6 . A 

dt dQ dt 

Turning back to § 37, we find the assumption that — 

small displacement of piston 
corresponding small displacement of crank-pin 

velocity of pis to n_ 
velocity of crank-pin 



D.C. of a Function of a Function. 



117 



It is easily seen that this is the same thing as — 

dp 
dp _ 



rdO 



dt^ 

dO 

r — 

dt 



or that — 



dp _ dp dO 

~dt " Jo'Yt 



In that section we avoided the general assumption by 
showing, from other considerations, that in that particular 
case the result held good. 

It is now quite easy to correct this investigation for obliquity 
of the connecting rod. It is clear that, corresponding to the 
position C of the crank-pin, the actual displacement of the 




Fig. 41. 

piston / is not OM, but ON, where CN is a circle with P 
as centre. 

Let angle MOC = $. 
MPC = </>. 
PC - /. 
OC = r. 



Then/ = ON = OM + MN 
or p — r cos 6 + (/ — / cos <£) 
Now CM = r sin 6 = / sin $ 



(i.) 



Therefore sin <£ = - sin 6 , 



Oi.) 



ii8 . Graphical Calcultis. 

Differentiating (i.) with respect to t (see note at end of 
§ 34), we have (see p. 112 (v.)) — 

dp . a dO , d<f> 

because the d.c. of /= o (see § 19). But from (ii.) this 

- riBltf VS-5J ■ • • (m) 

We also have from (ii.) — 

r cos 0— = I cos <£ — 

ata /'cos# dO 

or -j- = • — 

«/ /cos<£ *// 

r cos rf0 



Vr--r*sm*0 dt 



since /cos <jf> = /A -sin 2 <f> = /\/ 1 - - sin 2 6 = V/ 2 -;- 2 sin 2 (9 

Substituting this value of -^ in equation (iii.) above, we 
obtain — 

.-0 



dp . A d0 ( r cos (9 

— = r sin # — — 

<# <# v *//■—>* sin 1 1 



which is the exact value of the velocity of the piston. If the 
connecting rod = n x length of crank, this becomes — 

dp . -dO [ cos 6 \ 

dt dt\ xV-sm 2 6> y 

This expression is rather complicated. It is simplified 
as follows : Sin 2 6 can never be > 1, whereas n 2 is always 
comparatively large, usually about 25. Hence V// 2 — sin 2 0, 



DC. of a Function of a Function. 119 

being very nearly = a/h*, is put = n. The maximum error 
in doing this is very small. It occurs when the crank is in 
the neighbourhood of 45 , at which point the actual error 
is about o'8 of 1 per cent, when n = 5. Making this 
approximation we have, since 2 sin cos 6 = cos 26 (see 
any book on trigonometry) — 

dp . V sin 2 6 

— = — V sin H 

dt 2n 

dO , . 

where V = r— = velocity of crank-pin 

= a constant 

Differentiating again with respect to the time, we obtain 
the acceleration — 

d 2 p v JB 2V JO 

•— = —V cos — + — cos 20— (see p. in (11.)) 

dr dt 2n dt 

It is well to lest the accuracy of equations of this kind by a process 
known as "taking dimensions." It is clear that in any equation what- 
ever all the terms must be of the same kind. It would, for instance, be 
absurd to have an equation such as the following :— 

2 — + 2 -= 5 seconds 



for a velocity can by no conceivable process be added numerically to an 
acceleration, much less can the sum of the two be equated to a time. 
Similarly, if our equation is correct, it is certain that all its terms, however 
obtained, must be of the same kind. This we can test by finding what are 
the dimensions of each of the terms (see note on p. 30, also p. 51). If 
these are not alike, it is very certain our equation must be wrong. Now, 
the left-hand side of the equation is obviously an acceleration, being the 
time-rate of variation of a velocity. This is also suggested by the form 
in which it is written, since d 2 p would naturally suggest a length. If this 
had been written {dp 2 ) we should have expected it to represent a small 
area, dt 2 represents, naturally, the square of a small element of time. 



120 Graphical Calculus. 

Hence -7? has for its dimensions — -# i.e. the dimensions of an accelera- 
di 2 sec. 

tion. Now consider the right-hand side. V is a velocity having for its 

ft. . . length 1 . 

dimensions ■ — . Cos has for its dimensions : 77 =-; or, in other 

sec. length 1 

words, has zero dimensions, or the dimensions of a simple number or 

dd . . angle . . 

"numeric." Now, — has for its dimensions — , but an angle has no 

at time 

dimensions, for the same reason that a cosine has none. Hence the total 

dimensions of the first term are ■ — — X - X = — — = an acceleration. 

sec. 1 sec. sec. 

Let the student work out for himself the dimensions of the second term 
on the right-hand side. In an ordinary algebraic equation, such as 
x 3 + 3-# 2 + • ■ . = o, each of the letters must be assumed to be of zero 
dimensions, i.e. to represent numerics (see note on p. 5). 

Both the expressions for the velocity and the acceleration 

dd 
are given in terms of 6 and — , and can therefore be found 

numerically by substitution. 

Examples. 

1 . By the method of Fig. 40 obtain the curves— 
(i.) y- sin(* 2 ). 

(ii.) y — sin (log x). 

(iii.) y = log (sin x). 

(iv.) y = log (log jt). 

(v.) y = log (cos e x ) 1 . 

(vi.) y — log {log (a + bx n )}. 

Differentiate them by (i.) multiplying together corresponding ordinates 
(as explained on p. no) of the respective derived curves; (ii.) by the 
method of Fig. 9 ; (iii.) algebraically, and plot the curve by calculation. 
Compare the results. 

2. Differentiate — 

(*•) 1°§ (V x —a + v x — b). Ans. — . 

2V(x-rt)(l-^) 

(ii.) l ax sin m rx. Ans. l ax sin m_1 rx (a sin rx + tnr cos rx). 

1 For this example and the next, the process must be a compound one. 
Thus : Find^ = cos e x by the method of Fig. 40, and y =log cos e* by a 
repetition of the process. Find by induction which derived curves must be 
taken for the factors of the result. 



B.C. of a Function of a Function. 12 1 

(iii.) x x . Take logs thus — 

Let y — x* 

therefore log y = x log x 

_.«. . . I dy . , x 

Differentiating - — - = log x + - 

ydx x 

d £=yQog*+i) 

= x* (log x + i) 
1 
,. I — tan x . 1 • \ \ 

(iv.) . (Arts. - (cos x + sin x).j 

sec # 

(v.) /**. Am. I* X x* log /X (1 + log *). Take logs twice in 
succession. 

(vi.) [log {log (log*)}]. Ans - 



x log * ■ log (log x) 

(vii.) V2 ax —x 2 . Am. ( . ~) 

\sl 2ax —x'y 

2X 2 

(viii.) tan- 1 ; . Am. — r 

v I — x 2 I + x 2 



3. Find the exact velocity and acceleration of a piston of an engine, 
given crank 8 inches, connecting rod 30 inches, revolutions 95 per minute, 
at angles 30 , 45 , 6o°, 90 , 120 , and 150 . 



CHAPTER IX. 

integration. 

§ 46. Examples of Integration. 

We have already explained, in Chapters II. and III., the real 
nature and nomenclature of the process called integration, 
and have obtained the integral of one function of x, viz. 
x n , which is — 

fx n dx = -±-x {n+1) 
J n-\- 1 

It has also been pointed out that, to effect any proposed 
integration, it is essential that we have a previous knowledge 
of the process of differentiation ; and it is only by working 
backwards from this knowledge that we can obtain an expres- 
sion for an integrated curve, though we can graphically find 
the curve itself independently of its equation. 

There are many simple integrals which we can write down 
at once if we know the corresponding proposition of the dif- 
ferential calculus ; but it should be clearly understood that 
there is no general method by which we can deduce the 
integral of a function from first principles in the same way 
as we have deduced the d.c. of various functions. The process 
of integration is essentially a tentative process depending on 
a previous knowledge of the differential calculus, just as 
the process of division in arithmetic is a tentative process 
depending on a previous knowledge of multiplication. It is 
impossible, therefore, to attain proficiency, or even facility, in 



I?itegratio?i. 123 

integration without a previous familiarity with the differential 
calculus. The expressions the integrals of which we can write 
down at once are the results of the differentiations explained 
in Chapters VI. and VII. Thus we have — 

ycos x dx = sin x + c 

which means precisely the same thing as — 

/(sin x + c) 



dx 



= cos x 



in much the same way as — = 3 means precisely the same 

4 

thing as 3 x 4 = 12. 
Again — 

f ( — sin x)dx = cos x + c 

which means the same thing as — 

^(cos x + e) 

— ' = - sin x 

dx 

This may also be written — 

ysin x dx - —cos x — c 

d( — cos — c) 

for — = sin x 

dx 

Also^rsin x dxdx, which means, as already pointed out — 

f (fsm x dx)dx =f ( — cos x + c)dx 

= — sin x + ex + e (see § 36 and p. 45). ] 
Also — 

fffsm x dxdxdx =/[/ {/( Sln xdx)dx}^dx 

=f[f {(-cos x + e)dx}y* 
s=y ( — sin x + ex + *)*& 

= cos „t + -x' + ex -\-f 
1 See note (n), p. 195. 



\ 






124 Graphical Calculus. 

Also we have I dx, which is usually shortened 

J vi— or 



into — 

dx 



j 



Vi— x l 
corresponding to — 



= sin -1 x or = cos" 1 x 



^(sin" 1 x) i ^(cos" 1 #) 

^c ~ Vi -x* *fo 

This result is sometimes confusing to the student. How 
is it that the same expression can have two different integrals ? 
To answer this we must refer to §§ 8, 16, 22, in which it was 
shown that in graphically integrating a curve we have to 
assume some arbitrary point to commence from, which point 
may be at any height above or below the base-line OX. At 
whatever point we start from in the same vertical line, we shall 
obtain the same shape of curve. If we draw two such curves 
starting at different points, any ordinate of one is greater than 
the corresponding ordinate of the other by a definite and con- 
stant amount. 

Now, bearing this in mind, let us look at Fig. 36, 
which shows the curve y = sin" 1 x and its derived curve 

y y = ± If we integrate the latter graphically, start- 

v 1— x 1 

ing at O, we shall, of course, obtain the curve y = sin -1 x\ 

but if we start at a point P at a distance — = 1*57 units lower 

down, we shall obtain the precisely similar curves = cos" 1 x 
as shown. In fact, the angle whose sine is x is greater by 

exactly -, or 90 than the angle whose cosine is x, for all 

2 

values of x. It will be evident from this example that a 

complete solution of the integral ^ is not sin -1 x or 

J v 1 — x 2 



Integration. 125 

cos -1 x, but it must be some function of x which will include 
both these functions and an infinite number of other similar 
ones, for we may start to draw our curve from any one of an 
infinite number of points on the vertical OY. Now, any curve 

, , . . r dx . 

which would answer to the description y = I + — : , i.e. 

J v 1 — x 2 

which might be obtained from the curve y y = + - by 

v 1— x 2 

graphical integration would be included in the equation 

y - sin -1 x -f- some constant. 

For different values of the constant we should get different 

curves, but all of exactly the same shape. If the constant were 

( -.- J we should obtain y = sin -1 x — -, which we have shown 

above to be the same thing as y = cos -1 x, whereas if the 
constant were o we should have y = sin -1 x. 

Now, in every case of an " indefinite integral," i.e. without 
any limits specified (see § 22), this unknown constant must be re- 
presented by a letter, though it is often omitted for convenience, 
unless more than one successive integration is required, when 
it must never be omitted (see Examples 1 and 2, p. 30 ; also 
p. 45). It will also be evident that we can in general find 
the exact value of this constant, if w T e know one point through 
which the integrated curve passes. But in the above case, if 
we know that x = o when y = o, we know that the constant 
must be either o or mr where n is an integer. If we know, 



^-& v 



in addition, that — at O = +1, we know that the constant 
ax 

is either o or 2/ztt, i.e. it x an even number, either of which 

would give us exactly the same result. If — at O is — 1, then 

the constant must be (2/2 + i)tt, i.e. it x any odd number. 
This is perfectly definite, for all odd numbers would give the 
same result. 



126 Graphical Calculus. 



§ 47. Example of Quadrature of Area. 

It has been already shown, both graphically (§ 10) and 
algebraically (§ 22), how the constant disappears if the 
integral is taken between definite limits. 

As an illustration, let us find the area of the curve y = sin x 
(Fig. 42) between the limits x = 1 and x = 2. It has been 
already shown (§ 21) that this area would be represented by — 

sin xdx - T — cos x + constant I 

This is the notation usually employed, the constant being 
represented by " C." For the sake of definiteness, let us 
assume a particular value for this constant, say 1^ units. Now, 
if this constant had been o, the integrated curve y - — cos x 
would have cut OY at a point where y = — cos o = — 1, 
as shown in the dotted line in Fig. 42 ; but since we have 
arbitrarily added ij to this value, the curve lies as shown, 
where OA = 0*5. Draw in the limiting ordinates p'P and 
/Q at distances of 1 and 2 units from OY. The upper 
curve is y = —cos x + i-j, and we know (§ 13) that gQ — /P 
in inches = number of square inches in area P^V'Q'- Now — 

^Q = —cos 2 radians -j- ij 

= -cos 114 39' + ii 

= 0*401 +• i*s = i'9o 
pV = —cos 1 radian + 1*5 

= -cos 57° 19'+ 1*5 

= -0-540 + i's = 0-96 

Hence — 

^Q-/P = (0*401 + i-5)-(-o'54o + 1-5) 
= 0-401 + 0-540 + 1*5 - 1-5 
= 0*941 sq. units 

The constant, it will be seen, disappears entirely in the 



Integration. 



127 



result, so that its absolute magnitude is a matter of no 
importance. 

We shall often find that the adoption of two different 
methods of integration will give us a different result for the 



< — X 




Fig. 42. 



same function. If, however, the work has been correct^ it will 
invariably be found, on examination, that the two curves re- 
presented by the two results are of exactly the same shape 
and are exactly alike in every particular^ except that one is 
higher than the other by a definite and constant quantity all 



128 Graphical Calculus. 

along its length, and that the one algebraical expression for 
value of y can be expressed as the sum of the other and a 
constant 

§ 48. Work done by Expanding Gas. 
Another function for which we found the d.c. was log x. 
The result was - (see § 40). 
Hence — 

log x 



Pdx = 

J X 



or, as we have seen the general integral to be, log x + c. 

This result is of great importance. It is constantly occur- 
ring in engineering problems. It furnishes, for instance, a 
solution of the question as to the amount of work done by 
compressed air or steam in expanding from one pressure or 
volume to another. 

Take the case of air. Boyle's law tells us that if air 
expands or contracts at a constant temperature, the pressure 
varies inversely as the volume, or, in other words, pv = con- 
stant. This constant can easily be calculated from the mass 
of air and the temperature. For 1 lb. of air at 32 Fahr. the 
value of the constant is 26,214, when the pressure is measured 
in pounds weight per square foot, and the volume in cubic 
feet. For half this quantity of air the constant is, of course, 
T 3> 10 7 > f° r at tne same pressure the volume is half what it 
was before, and therefore the product pv has half its previous 
value. In the same way, since the volume varies directly as 
the absolute temperature (i.e. temperature Fahr. -j- 46 1° nearly), 
pressure being constant, this product must vary according to 
the same law, as may easily be seen by imagining the pressure 
kept constant, while the temperature, and therefore the volume, 
varies. The constant may in all cases be calculated by finding 
the value of the expression, 53*2 x m X t, where m = mass of 
gas in pounds, r = absolute temperature. 



Integration. 1 29 

Now. if all these values of/ and v for a given mass of gas 
at a given constant temperature be plotted on a curve (pres- 
sures-vertical), the resulting curve will be a rectangular hyper- 

, , . constant '. . ... .. 

bola, whose equation is p — . It is the "indicator 

v 

card" of the expansion, and it is shown in all works on the 
steam-engine, in a similar way to that adopted in § 14, that 
the area under the curve between any two ordinates repre- 
sents the amount of work done during the expansion between 
the corresponding volumes. 

The chief difficulty in understanding the working of 
these problems is that of units, which will continually harass 
the student till he masters it once for all. He must here 
imagine the curve drawn to full inch scale, i.e. 1 inch vertical 
= 1 lb. per sq. foot, 1 inch horizontal = 1 cub. foot. Under 
these circumstances, 1 sq. inch on the diagram represents 

-— ' x ft. 3 = 1 ft. -lb. If the scale had been 1 inch vertical 

= 1000 lb., and 1 inch horizontal = 10 ft. 3 , an area of 1 sq. 

inch on the diagram would have represented 1000 — - 2 x 10 ft. 3 

= 1 o,ooo ft.-lbs. 

This may be taken as an example of the general method of finding the 
scale in which an area, such as an indicator diagram, measures a quantity, 
whose value we require. The rule is, consider what quantity would be 
represented by a square figure one inch long and one inch high. 

Taking, then, the full-size diagram, we have its equation — 

26214 
p = 4 

v 

for 1 lb. of air at 32 Fahr. 

Its integrated curve, as we have seen, is a curve of loga- 
rithms, each of whose ordinates is X 26214 (see § 40), whose 
equation is therefore — 

y = 2 6, 2 1 4 log v -f- e 

K 



130 Graphical Calculus. 

The area, then, of the lower curve between any two ordinates 
(say, where the volumes are 5 cubic feet and 9 cubic feet) is 
the difference between the two corresponding ordinates of the 
upper curve — 

= 26,214 log 9 + c — 26,214 log 5 — c 

= 26,214 (log 9 - log 5) 

= 26,214 log f = 26,214 X 0-588 

= i5>4i3'8 

This is the area of the curve, and since each square inch 
represents 1 ft.-lb., the total work done is 15,4138 ft.-lbs. If 
any constant other than 26,214 had been given with the same 
ratio of expansion, this constant, instead of 26,214, would 
have been multiplied by log f. 

Thus, suppose in an air-compressor, diameter of cylinder 
= 10 inches, stroke = 2 feet; required the work done per 
stroke in compressing air isothermally up to 6 atmospheres. 

Here volume of air compressed per stroke = io 2 x 07854 x 24 

= 1884*96 cub. in. 

The corresponding pressure is that of the atmosphere, viz. 
147 lbs. per square inch; 

The constant therefore = 1885 x 147 = 27709-5 



ft. 3 and — '- to in. 3 and .—4 If we plot this expansion curve 



Notice very carefully the effect of altering the units from 

lb. . 3 lb. 

— : to in/ and . — x 
ft. in. 

in these units, one square inch of the diagram will represent 

1 _ 1 x 1 in. 3 = 1 in. -lb. Therefore we must divide the area 
in. 2 

by 1 2 to get foot-pounds. The result is — 

. ,, 27709-5 X 17918. „ _ 
27709-5 log { in.-lbs. = n ^ u — ft-lbs. 1 

1 Part of this work, viz. — — ft.-lbs., has been done by the 

12 

atmosphere which presses on the suction side of the piston. 



Integration. 131 

The student should in every case pay great attention to 

the units in which he is working, otherwise he will find himself 

hopelessly confused. For instance, if in the above case he 

lbs. 
had taken pressures in -r— 2 and volumes in ft. 3 , then one square 

inch of diagram would have represented -7—7- X ft. 3 = 1728 

in.-lbs. 

It may be noted that in an actual air-compressor the work 
would have been greatly in excess of this, because a large 
amount of heat is developed in the air by the process of 
compression, which increases the pressure, and therefore also 
the work done. 

If the compression is effected without any heat being lost, 

as will very nearly be the case if it is done very rapidly, it may 

be shown that — 

^1-408 _. cons tant 

The constant here, also, will have to be calculated either 
from known simultaneous values of the pressure and volume, 
with the help of a table of logarithms, or from the temperature 
and volume and mass. In this latter case the, constant = m x 
53*2 x r x ^°' 408 = c> suppose. Here, as before, the work done 
is — 



but / = ^JS 



/: 



pdv 1 

Vi 



Cvo cdv Cvo Cv 2 

hence integral = — - = cv n dv = c v n dv 

J v x v im J v x J v 1 

where n = — 1*408 

1 Here it will be noticed that it is impossible to integrate this as it 
stands, because the expression to be integrated, viz. p, does not contain v 
at all, and dv tells us that the expression has to be integrated with respect 
to v. Hence our object is to change p into an expression containing no 
other variable except v. This we must do by "eliminating" p between 
the two equations, or, in other words, substituting for p an equal value in 

c 
terms of v* ue.-r^ 



132 Graphical Calculus. 

Hence the work done is — 



n + 1 



r,n + 1 



or, substituting p x v^' m for c and — 1*408 for n- 



PiV. 



v 2 

1-408 



1 ^-0408 I 

— 0*408 



Hence, substituting v 2 and v x in turn for v, and subtracting, 
we obtain — 

^-riPiVi'"* x zv -0 ' 408 - A^i 1 ' 408 x *>r ' 408 ) 

0*408 

but since we know that A^i 1 ' 408 = A 7 ^ 1 ' 408 * we can write this — 

which represents the work done. 

§ 49. Integrals to be Learnt. 

The following integrals must be learnt by heart. The 
corresponding differentials are also given in § 42. 



I x n dx = —^-x {n+1) 
J n + 1 

Jf = !<** 

a x dx - : 

j sin #dfo = — cos x 
Jcos^sin* 



Integration. 133 

dx 



= tan x 



COS X 

dx 

— - = cot X 

sirr # 

LI t- V # _ cV 

sin"" 1 - 



\/tf 2 — #* ^ 

^ 1 , ^ 

= -tan" 1 - 



s-. 

/: 

/ 

J a 2 + x 2 a '"' a 

C dx 1 x 

■ = - sec -1 — 

J x,Jx 2 -a 2 a a 

These are the most important elementary integrals. The 
integration of any expression which can be integrated is 
effected by transforming it by processes which will be shortly 
explained, into forms of which the integral is known. 

Many simple cases can be so transformed at once. For 
instance, required — 

rdx_ 

J V f x 
This can be written — 

f x~^dx 
This is evidently an example of the x n integral given above, 

where n = — £. Since we know that x n dx = x n + \vre 

J n + i 

have evidently — - 



X 



X~idx = X~i + l = 2X* 

2 + I 



= 2\l X 

To test the accuracy of this, let us differentiate the latter 
expression — 

d( 2 Vx) _ d(ix\) 



dx dx 

1 

V ' x 



= 2XlX 3G- 1 ) 



134 Graphical Calculus. 

or again, required — 

C /9t C t 

; X r dx 



C dx Ci 

J (p4 )p r 



Here — - = n 



ii !— i 

hence required integral = — x ■ x x r 

- Q 

fi r I-- 

r 

Again, requiredy cos mxdx — 

Here, if we try sin mx as the resulting integral, we shall find, 

on differentiating it — 

<^(sin mx) 

= m cos mx 

dx 

It is evident, therefore, that sin mx is m times too great ; 
therefore, instead of taking sin mx, we evidently ought to have 

had — sin mx, which on differentiating gives cos mx. This is, 

therefore, the correct result. 

It is to be noticed that functions of (x + a), where a is 

any constant, can usually be treated exactly like the same func- 

^ d sin (x + a) , N 

tion of x. For instance, = cos (x + a), and 

dx 

dlog (x-\-a) i r , „ , , _ . , 

= . for the d.c. of (x + a) with respect to 

dx x + a 

x = i (see § 43). This principle does not, of course, extend 

to such expressions as log (x°- + a 2 ), whose d.c. is evidently 

1 

—. 2 X 2X. 

x l + cr 

Similarly, we are continually dependent on our previous 
experience of the differential calculus to enable us to effect 
any proposed integration, and it is thus evident that we must 
have the d.c.'s of the elementary functions at our finger-ends 
before we can hope to attain facility in integration. 



Integration. 135 

/dx 
~ r — — ^ Even 
a" + x" 

in simple case of this kind we are at once hopelessly lost, unless 

we happen to know that -y- ( tan" 1 - J = -»—. — r > 
rr ax \ a J a~ -{- x J 

x 1 

Now, if we tried tan -1 -, to see if it would produce -r— — ; 
a or 4- -ar 

on differentiation we should find it was a times too large ; hence 

1 ,# 

the correct result is - tan" 1 - 

a a 

Such a procedure is, no doubt, extremely unsatisfying to the 

student; at the same time it is the only one that is open 

to him, and he must be content to make the best of it by 

continued practice. He will find even the elementary integrals 

and a few easy applications to be of great service to him in 

practical work. 



Examples. 



Integrate a 3 *, a 2x x e x 9 sin x cos x, ( — ; — : + : ) 

& \ 1 + sin x 1 — sin x J, 

I I \ / I \ ( {X - ff + 2pX \ 

JlT+a^'b ' \/a - x + bf \x - sin xf \ {x 2 +f) 2 ) 



,3, -, : ( x $*Jx — a\ 



CHAPTER X. 

methods of integration. 

§ 50. Integration by Expansion. 

Many expressions can be integrated by expanding them into 
separate terms by some algebraical or trigonometrical process. 
This method should always be tried before any other. 

Take, for instance, f (a 2 + x 2 fdx. On expansion, this 
becomes — 

J (a* + 3 a 4 x 2 + 3^V + o?)dx 

This expression, as we have shown (§ 29) — 

= J a G dx + f 3 a 4 x 2 dx + f 3 a 2 x i dx + f x'dx 
= a 6 x + a'x" + f «V + \ x 7 
Again — 



f dx _ fj_ f 1 1 "\ , 

\x l — a 2 \ 2a\ x -a x + a / 



1 



dx 1 dx 



2<x \ x — a 2a I x-\- a 

2a oV ' 2a oV ' 

1 _ x — a 
-log 



2a x-\-a 

This is a very important result. 

Again — 

fs'm 2 xdx= \/2 sin 2 x 



Methods of Integration. 137 

= \ f(i — cos 2x)dx = \fdx — \fo.o>% 2xdx 
= \x — ^-sin 2x 

Examples. — Integrate (x + a) 2 , (px + q) 3 , (x + a)(x — a), 

~ 8 IT. 

-, (cos x -f- sin xf, -— —- — — -. (Split this deno- 



x — a ' x 1 -{- $x -\- & x" 1 — 6 

minator into (x + V<5)(# — V6).) 

It is not necessary or desirable to give the answers. It is 
always possible to find whether the result is correct by dif- 
ferentiating the result obtained. The d.c. of the answer should, 
of course, be the same as the function to be integrated. 



§ Si- 

If it is found impossible to reduce the proposed expres- 
sion into a series of simple integrable forms, the next thing to 
be done is to try whether it is possible to write it in the form 
of two factors, of which one is the d.c. .of the other, or of some 
power or root of the other. If so, the expression can be 
immediately integrated. 

Thus to findy(V + cPx^dx. Here we see that we can 
write this in the form — 

fx*Jx 2 + a 2 dx = \f2x X (x 2 + a 2 )idx 

Now, 2x is the differential coefficient of x 2 + a 2 . 
Now, obviously, if we differentiate (x 2 + a 2 )h + i, we shall 
obtain (§ 43)— 

%(x* + a 2 )h x 2x = 3 x\^x Y + a 2 

This is three times too large. Hence the required function 
is \(x 2 + a 2 )h 

Again, required f (px 2 + 2qx + ?f(px -f q)dx. This 
becomes — 

\f(px 2 -f 2qx + r)\(2px + 2q)dx 



138 Graphical Calculus. 

By a similar process, we obtain — 



/\ y\/ / * v ~t~ *¥>* ~x~ f 

A similar case is- 



i x |(^ +2F + ; -)I 



ax + b _ . / 2ax + 2b 

ax = -k I — 

ax' + 20X + c J ax' + 2bx + ^ 



Here the numerator is the d.c. of the denominator. When 
this is the case, we at once write down the integral 
log (ax 1 + 2 foe + c). The student will see the reason for this 
if he differentiates this expression. 

Examples. — Integrate —^— ( this expression = (log x)x- J , 
x \ jc •/ 

1 # jc / sin # \ 

tan # I = - J , sec x tan ^ 



# log x px 1 -\-q s/px 1 + q* ^ cos a; 

sin x \ sin 2# 



r- 



cos 2 X J ' COS 2 X' 



§52. Integration by Substitution. 

The next method to be tried is more difficult to under- 
stand. It consists in changing the variable from x to some 
.other, usually z, by substituting z for some function of x con- 
tained in the expression to be integrated. By this means, as 
will presently be shown, we can often, by judicious substitution, 
reduce a complicated expression in x to a simple one in z. 
This is really treating the expression exactly as in the last 
case, but we shall be able to deal by this method with more 
difficult examples. 

Take a simple case for the purposes of illustration. Find 

/ x\ ^ 
the area of the curve y = ( 1 + - ) between the ordinates 

x = 0*5 and x = 2*0. As we have seen, this area is represented 

f 2 ' f x\i 
by [i+-)dx. Now, suppose we substitute z instead 



Methods of Integration. 139 

fx = 2'0 
of the expression in the brackets, we obtain zldx. 

J * = 0-5 
At first sight, the meaning of this is far from clear. The 
student will have seen that before we attempt to integrate any 
expression we must first of all get it into some form in which 
functions of one variable only are present, otherwise we 
cannot be sure of what we are doing. Since we cannot 
integrate this expression with respect to x, we are going to 

make z or ( 1 + - ] the independent variable, to see whether 

it is any easier to integrate in that way. To do this we must 
completely change every x in the expression into the corre- 
sponding value in terms of z by means of the known relation 
between z and x. In doing this we must not omit to change 
the dx into some multiple of dz. In order to explain the 

method geometrically, draw the curve y = f 1 + - J . The 

easiest way of doing this is as follows : — 

Draw a curve (Fig. 43) showing the relation between x and z, 

x 
or 1 H — , for all values of x between the given limits. For 

convenience of reference, call the ordinate of this curve z. 

This curve, which is shown at (a) in the diagram, has for its 

x 
equation z - 1 + -. Now the ordinates of our given curve 

2 

y - ( 1 + - J are the f power of the ordinates of curve (a). 

Transfer the ordinates of curve (a) to another horizontal base 
6 X 6 as shown, and on this base draw the curve y = zl by 
calculating the ordinates with a table of logs or a slide rule, 
and erect ordinates to it from all the points / 2 , q 2 , etc. 1 This 
curve consists of two branches (as shown) with a "cusp" at 
the origin (see Fig. 20), i.e. the tangent at the origin touches 
two branches of the curve at the origin. Next draw another 
base OgX as shown, and divide it exactly as O a X a is divided, and 

1 See note (12), p. 195. 



[40 



Graphical Calculus. 



erect ordinates from each of the points of division. Transfer 
each of the ordinates of curve (b) to the corresponding ordinate of 
this new curve as shown. Draw a smooth curve through each 
of the points so found (much of the actual construction is left 
out in the figure for the sake of clearness). This curve is easily 



seen to be the given curve y - 



2 



Draw in the 



limiting ordinates ,# =0*5 and x = 2*0, and the corresponding 
ordinates of curve (b). It is clear that, although we cannot 
at once find the area of curve (c), yet we can at once find 




T X 



Fig. 43. 

that of curve (b) between the corresponding ordinates, for it is 
fzidz, taken between proper limits, these limits being the 

x 
values of 1 H — when x is 0*5 and 2*0 respectively, which values 

are 1*25 and 2*0. 

Our object, then, is to find a relation between the area of 
curve (c) and that of (b). Consider any shaded element of 
area of (c), and the corresponding elements of (b) and (a). 
These elements of (b) and (c) are the same height, and their 



areas are directly as their breadths, and clearly 



P 2 M 2 
PM P^lx 



dz 

dx 



dz 



Now — = \ always, that is to say, dz = \dx. 



Hence the 



Methods of Integration. 14 1 

area of curve (b) between the ordinates 1*25 and 2*0 = 1 area 

of given curve (c) between the ordinates 0*5 and 2*0 for each 

element of (b) = i corresponding element of (c). The reason- 

dz 
ing would have been exactly the same if — had not been 

constant, as in the next example, for instance. The above 
shows the geometrical meaning of the following reasoning, 
which is that given in most text-books : — 

To find I f 1 + X - ) dx. 

Let 1 + - = z. 
2 

Then \dx = dz 

dx = 2dz 

Hence I f 1 + - J dx =■ f 2z\dz 

= 2 X \zl = \zl 

x\% 

Take another example (the student should not fail to 

xdx 



draw the curves in this example and the next) : ~. 



a 1 + x 1 



Put \/a 2 f x* = z 1 






Then a 2 + x 2 = 


z 2 




And therefore 2 xdx = 


2zdz i or dx 


- -dz 

X 


f xdx 

TTpnpp I — 


z _ 
fx . - .dz 
1 x 

J z 


-fdz = 


J Va 2 + x 2 


— 


*/a 2 -f- x 2 




Again, if we desire to find 1 


dx 
\^x 2 + a 2 





142 Graphical Calculus. 

Assume >J x 2 + a 2 = z — x. 

Then x 2 + a 2 = z 2 — 2zx + x 2 



dz dz 

2Z~~ = 2Z + 2X — - 

dx dx 



or 



dx z — x 
dz dx 



Hence 



or — = 

z z — x 

dx f dx 



Vx 2 + a 2 ) z " x J z 

= log z = log (x + s f x 2 + a 2 ) 



The student can only hope to learn what substitution will 
be required in any given case by continued practice. He is 
referred to a larger book for further examples. 

Examples. — Integrate , ; — , — -, —. 

Vx + a (^+3K (*-¥ 

x — a ax + b 1 1 

JTx-cY + P Jax 2 + 2bx + c Vx 2 + 4 V\x + af + 16 



§ 53. Integration by Parts. 

Another method of great importance is known as " inte- 
gration by parts," which presents considerable difficulty to the 
beginner, because of the large number — eight — of different 
functions which' have to be simultaneously borne in mind. 
The process is, in reality, the opposite of that explained in 
§ 30 for differentiating a product. 

Before commencing the following explanation, the student 
should carefully read § 30. That article showed how to find 
the d.c. of an expression represented by curve (3), which 
was the product of two other expressions represented by (1) 
and (2). It was there seen that if — 



Methods of Integration. 143 

(4) be the first derived of (1) 

(5) be the first derived of (2) 

(6) be the product of (2) and (4) 

(7) be the product of (1) and (5) 

(8) be the sum of (6) and (7) 
Then (8) is the first derived of (3) 

But suppose we had been given curves (6), (2), and (4) 
in their correct places, and had been required to complete 
Fig. 44, we should evidently have proceeded as follows — 

(a) integrated (4), thereby producing (1) 

(b) multiplied together (1) and (2), producing (3) 

(c) differentiated (2), producing (5) 

(d) multiplied together (1) and (5), producing (7) 

(e) addedtogether (6) and (7), producing (8) 

Now, it was shown on p. 75, that since ordinate of (6) 
+ ordinate of (7) = ordinate of (8), therefore area of (6) 
+ area of (7) = area of (8) (§ 29, p. 59), all areas being 
taken between corresponding ordinates. 

But area of (8) is represented by the difference of corre- 
sponding ordinates of (3), since (8) is the first derived of (3) 

(§ 13). 

Hence we have — 

(/) area of (6) = ordinate of (3) — area of (7) 

Now, suppose that, instead of the curves (6), (2), and (4), 
we had given their equations, and had been required to find the 
area of (6) (or, in other words, its integral), we might proceed 
exactly as at (#), (£), (c), (d) above, by writing down the equa- 
tions to the curves instead of drawing the curves themselves, and 
by the same processes as there described finding the equations 
(3) and (7). Then by the use of relation (/) we can make 
the integral of (6) depend on the integral of (7). The use of 
the process consists in this, that sometimes (7) is an easier 
expression to integrate than (6). If, in any given case, it is 



144 



Graphical Calculus. 



not so, then the process is of no assistance, and some other 
must be tried. 




Fig. 44. 



An example will make the meaning of this clear. 
Required the integral of x n log x, an expression which we 
cannot integrate immediately. 



Methods of Integration. 145 

Here equation (6) is y - x n log x 
equation (4) is y = x n 
equation (2) is jy = log x 

The product of any ordinate of (4) with the corresponding 
one of (2) is then equal to that of (6). Now, obviously, as at 

above, (1), being the integral of (4), must be y = — — — x' n + 1) . 
(b) . . . (3), being (1) X (2), is 7 = ^^ x **+« 



x 



(c) . . . (5) being the first derived of (2) is y = - 

(d) . . . (7), being (i) x (5), isj =~ f ^r I x * 



x" 



(n + 1) 



Area of (6). Ordinate of (3). Area of (7). 

x n dx 



(/) We have x n log x dx = -^i£. x 'n + i) _ 

J n + 1 Ji 



(n + 1) 

_AT w+1) l0g.X X n + 1) 

{n +if " (« + i) 2 

Exactly the same method may be applied to findy*^ sin x, 
which is curve (6) in Fig. 44. This is left as an example for 
the student. 

In working examples, the beginner will save himself a great 
deal of confusion if he writes the symbols down in the same 
relative positions as the corresponding curves in Fig. 44, till he 
has become thoroughly familiar with the process. 



Thus —cos x 
x 

— X COS X 



sin x 

1 



x sin x 

— cos X 

(x sin x — cos x) 

L 



146 



Grapliical Calculus. 



The symbols corresponding to the curves in the diagram, 
the order of writing down is as follows : (6), (4), (2), (1), 

(5), (3), (7). (8) may be left out, 

as it is not needed. 

Another geometrical illustration which 
the student should completely analyze 
for himself is as follows. In Fig. 45, 

PABQ represents the area I udv, and the 
J a 

[ d 
area CPQD represents the area I vdu 

when c and d are respectively the values 
which v assumes when u has the values a and b. Here it is clear that 
area PABQ = area CPABQDC - area CPQD, or fudv = uv - fvdu, 
all taken between corresponding limits, u and v are both supposed to 
be dependent on the independent variable x, which does not appear in 
the diagram (see § 45 and note to § 23 on p. 49). Hence the above equa- 
tion is really a shortened form of 




/ u—dx — uv — I —dx 



dx 



Of course, it is only by trial that the student can discover 
whether or not the process is of any use to him ; that is to say, 
whether or not equation (7) is any simpler to integrate than 
curve (6) ; if not, of course the process is useless. It is to be 
noticed that integration by parts can almost always be tried in 
two ways, viz. by putting each factor in turn in position (4). 
Thus, in this instance, if we had written the process thus — 



if 

sin x 

ix 2 sin x 



x 
cos x 



x sin x 

1 . 



(x sin x -j- \x 2 cos x) 
cos x is no easier to integrate than 



it is evident that 
x sin x. 

Take another case, f log xdx. Here there is, apparently, 
only one factor, but we can use 1 for the other factor, and 
proceed thus— 



Methods of Integration. 



H7 



x 

log X 
x log X 



I 

I 

X 



log X 

I 



Hence/* log xdx — x log x — fidx 

= x log X — X 

It is sometimes possible to integrate an expression by in- 
tegrating by parts twice in succession. Care must be taken 
which expression is placed in position (4) in the second opera- 
tion, otherwise the only result will be to reproduce the original 
expression. Several of the following examples must be treated 
in this way. 

Examples. 

Integrate x cos px, xex, x 3 e* t x sin x cos x, xn log x, ;m(log x) 2 , 
eux sin bx. (Integrate by parts twice, and reduce the required integral by 
the principles of simultaneous equations from the equations so obtained.) 

Sin aJx (put J x = z, as in § 52 ; then integrate by parts), (log x) 2 , 
x z log X 2 , 



CHAPTER XI. 

MISCELLANEOUS APPLICATIONS OF DIFFERENTIATION. 

§ 54. Maxima and Minima. 

In § 10 (i.), (ii.), (iii.), which should be re-read, it was shown 
that the height of a derived curve corresponding to a maxi- 
mum or minimum on the primary (which terms were there 
explained) was o, or at that point the primary curve was of no 
slope. Conversely, if at any point the first derived cuts OX, 
then the corresponding point of the primary is a maximum or 
minimum. It was also shown in (iv.) that we can distinguish 
between a maximum and a minimum by the direction of slope 
of the first derived curve. In other words, if the second 
derived curve is at that point above the axis of X, then the 
point on the primary is a minimum. If at that point the 
height of the second derived curve is negative, then the point 
on the primary is a maximum. 

These principles are of great use in practice from the ease 
with which we can find algebraical expressions for the height of 
a derived curve. If we have an expression involving x which, 
as x increases, first increases and then diminishes, we can 
find that value of x at which the function has attained its 
maximum value, by finding that value of x which makes the 
first derived function = o. In other words, differentiate the 
function and equate the d.c. to zero, thus giving an equation 
to find x (see Example in § 2). Differentiating the d.c. thus 
found with respect to the same variable, we evidently 



Miscellaneous Applications of Differentiation, 149 

obtain an expression for the height of the second derived 
curve at any point. Substitute in this expression the value 
of x found by equating the first derived function to zero. If 
the result is negative, the height of the primary is at the corre- 
sponding point a maximum ; and if positive, a minimum. 

The successive derived curves of sin x form a very in- 
telligible illustration of this. Suppose we wish to find at what 
points the value of sin x is a maximum or minimum. The 
height of the first derived curve is clearly (Fig. 46) given 
by cos x for all values of x. Now, equating this to zero, we 




Fig. 46. 



find a value such as O r S r of x for which the first derived curve 
cuts the axis of x. 

These values are found from the equation cos x = o. 

Hence x = - — — , etc., each of which corresponds to 
222 

either a maximum or a minimum. 

To find which is which, differentiate cos x, giving — sin x. 

Now substitute the above values of x in the expression 
— sin x, and see whether the result is a positive or negative 
value. 



ISO 



Graphical Calculus. 



— sin- = — i, showing that at a point distant - from OY 
2 2 



• 3 71 " , 

— sin — =+ i 

2 

— sin — = — i 



the primary curve attains a maximum. 
„ „ minimum. 



maximum, 



etc., etc. 

These can easily be verified in the diagram. 
The way in which this principle can be applied in practical 
case will be seen by considering an example. Suppose a man 
running along a footpath AO (Fig. 47) wishes to reach a house 
at H, in the middle of a ploughed field, in the shortest possible 
time. Suppose that he can run on the footpath at the rate of 
15 feet per second, but only at 10 feet per second in the 
ploughed field. What will be his quickest route across the 
field ? If he ran across AH, he would have the shortest possible 

distance to go, but his speed 
would be only 10 feet per second 
instead of 15 along the footpath, 
whereas if he went to O along 
the footpath he would have a 
greater distance to go, but his 
average speed would be greater 
than in the other case. It is 
clear that the shortest time 
would be occupied by taking 
some intermediate route, A/H, 
where he would partly utilize his 
superior speed along the footpath, and partly the advantage of 
cutting off the corner /OH. It is our object to determine 
the position of the point/. 

Suppose the distances are OA = 500 feet, OH = 150 feet. 
Let Op (the distance to be found) = x. 

Then distance to be run in the field = slop + (150) 2 




Fig. 47- 



Miscellaneous Applications of Differentiation. 1 5 1 



Time occupied = -*— \J±iL seconds 

10 



Distance to be run along road = 500 — x 

COO — j$ 

Time occupied = : — seconds 

^ i. w ■ j *J x l + ( I 5 ) 2 . 5°° ~ x 

Total time occupied = y = + 

10 15 

We have now to find the minimum value of y. 

Plot various values of y along OA as base, and draw a curve 
through the points so found. Thus <?Q represents the time 
occupied if the man leaves the path at q. Now, this curve is 
obviously parallel to OA at the point where the derived curve 
(a line not shown in the figure) cuts 0\X\ This value oix will 
be found by equating the d.c. of y to zero (see example in § 2). 

Simplifying the equation, we have — 

_ 3VV + (i5°) 2 + Jo 00 - 2X 

Now, since we only wish to find for what value of x this is a 
minimum, it is clear that we may simply consider the numerator, 
for if the numerator is a minimum, the whole fraction will be a 
minimum. (If the denominator contained x, or anything de- 
pendent on x, we could not legitimately disregard it.) Similarly, 
we may disregard the 1000, since it is always the same, what- 
ever be the value of x, and our problem becomes to find for 
what value of x, $\lx 2 + (150) 2 — 2x is a minimum. 

Exercise. — Let the student determine for himself the 
graphical interpretation of the process of disregarding these 
constants. 

Differentiating this, we obtain (§ 43) — 

3 

X 2X —2 



2 V f X 2 + (150) 2 

This represents a multiple of the tangent of slope of the upper 



152 Graphical Calculus. 

curve, and must therefore = o at a point corresponding to P, 
Therefore, to find x, we have the equation — 

3 * - 2 = o 



si x l + (150) 2 



or, simplifying- 



Sx 2 = 4 X (150) 2 
whence x 2 = -f 150 2 
therefore x = 134 yards about 

the negative value being obviously inadmissible. 

Another familiar example shortly worked out may serve 
to further illustrate the process. 

According to the post-office regulations for the size of 
parcels which may be sent by parcels post, the length of 
parcel + girth must not be greater than 6 feet. Required 
the greatest volume which can be sent. 

Let x feet be the girth ; then 6 — x = length. 

Now, with any given perimeter, the figure containing the 

greatest area is well known to be a circle. 

x 2 
The area of a circle of girth x = — 

47T 

x 1 
;. volume of a parcel of girth x and length (6 — x) = — (6 — x) 

4?r 

Therefore the question becomes for what value of x has 
6x 2 — x 3 a maximum value. 

Differentiating and equating to zero, we have — 

i2# — 2> x<2 = ° 

Neglecting the solution x = o, which obviously gives a 
minimum, we have — 

x = 4 

The volume is therefore - = 2-55 cubic feet 

7T JJ 



Miscellaneous Applications of Differentiation. 153 

Examining these three examples, the student will see that 
the rule is — Express the quantity of which we have to find 
the maximum or minimum value in terms of one variable ; 
differentiate the expression with respect to that variable ; 
equate to zero, and solve the resulting equation. The solu- 
tion gives the value of the variable for which the expression 
has either a maximum or a minimum value. If it is not 
apparent on inspection whether the value so found gives a 
maximum or minimum, differentiate the first differential 
coefficient, and substitute in the expression thus found the 
value in question. If this gives a negative result, the value 
gives a maximum; and if a positive result, a minimum (see 

§ 10, 4). 

Exai7iple 1. — A man weighs 160 lbs.; he attaches a rope 
to the top of a post 20 feet high, with the object of pulling it 
over : what is the greatest bending moment he can produce 
at the base of the post, assuming that the pull he can exert 
on the rope varies as the sine of the angle which the rope 
makes with the ground? Ans. 1600 lbs. feet. 1 

Example 2. — Find the minimum weight of a cylindrical 
boiler made of i" plate necessary to hold 200 cubic feet of 
water. Neglect overlap of plates. One cubic inch of iron 
weighs 0-28 lb. Ans. 3850 lbs. (about). 

Example 3. — One leg of a pair of compasses is held 
vertical with its point stuck in a board, and the compasses 
are rotated about this leg as axis : find what angle the other 
leg must make with the vertical, in order that the bending 
moment tending to open the compasses may be a maximum, 
given that the legs are uniform, each 5" long and each 



°'$g± V ^- + 12 - 5 CO 4 

weighing i oz. Ans. cos 6 = ^ 2 where 



5(o' 



g = 32*2, and co = angular velocity. 

1 See note (13), p. 195. 



154 GrapJiical Calculus. 

§55. Indeterminate Forms. 

It sometimes happens that we have to find the value of 
some function of x which cannot be obtained by simple 
substitution, because on giving x the required value the 

function assumes the form ~, o°, o x &o , or some such form. 

o 

In these cases the application of the principles of the 

differential calculus enables us to effect a very simple solution. 

For instance, we might have to find the value of — 

X 3 — I 2X + 9 

when x == 3- 



x 3 — 4X 2 — $x-\- 24 



o 



It is easy to see that this fraction assumes the form 

o 

when x = 3, and we can therefore not determine its value 

by simple substitution. Assume that the two curves APB, 

CPD represent the values of the numerator and denominator 

respectively of any fraction ( -^—— ) , which assumes the form - 

\A{x)J o 

when x = OP. Now suppose, for the sake of definiteness, 

dy 
that we have found — for the curve APB to have the value 
ax 

i*5 at the point P, and for the curve CPD the value 2-5 at 

the same point. Draw the tangents ST and RQ at the point 

P. Take two verticals TQM and SRN near to P, but on 

, MT . 
opposite sides of it. It is clear that rrrzz is constant wherever 

we take T on the line ST. Now, when very near the point 
P, the points T and Q (as has been frequently explained) 
are respectively on the curves representing the value of the 
numerator and denominator of the given fraction. Now, when 
T has travelled through P to an extremely small distance 
the other side of P, it is clear that the value of the ratio 
has not altered either in sign or magnitude, because the 



Miscellaneous Applications of Differentiation. 1 5 5 



constant whatever be the sign and magnitude 



numerator and denominator have both changed sign, and there- 
fore their ratio has the same sign as before. Thus we have 
MT _ NS 
MQ " NR 

of the values PM, PN. It is therefore true when they are 
" infinitely small." Now, it is impossible to imagine what 
happens at the point P in just the same way as it is impos- 
sible to imagine an infinitely great or an infinitely small 






D - / 




B 
X 


/ P 

\ 7 1 

X^ _F' 


0' 


/f" 



Fig. 48. 



Fig. 49. 

quantity. Assuming the curve representing the value of the 
ratio at all points along OX is continuous, i.e. does not make 
a sudden vertical jump at the point P — and there is no reason 
for supposing it would — we say that at the point P it has the 
same value as it has at a point infinitely near to P. We 
therefore express the fact that, however small PM is, the 

dA(x) 
KF 



value of the fraction — — = x 

MT KE 



dx 
dx 



, by saying that the 



156 Graphical Calculus. 

value of the fraction at the point P = ratio of the values of 

the d.c.'s at that point. 

If the curves AB, CD are of the form shown at APB, 

df(x) df.(x) 

CPD in Fig. 49, it will happen that - n ^- i and ■ z ~ are 

dx ax • 

both equal to zero, in which case, by the same reasoning as 
before, the ratio will be that of the values of the second derived 
functions at the point. Again, one curve may be of the form 
APB, and the other of the form EPF, in which case the ratio 
will be = o or oc . This will be shown by one derived function 
vanishing, while the other is finite. 

Again, if the curves, in addition to having no slope at P, 
have a point of inflection at that point, the second derived 
functions will vanish at that point, and the ratio required will 
be that of the heights of the third derived curves at the point. 
In general the required ratio will be that of the first pair of 
derived functions which do not both vanish at the point. 

Examples. — Find the value of — 

tan x — sec x + 1 . , . . 

(i.) when x ~ o. (A /is. 1.) 

tan x — sec x + 1 

Iqct x 

(ii.) — ^— when x - 1. (A/is. 1.) 
x— 1 

(hi.) — when x = o. (Ans. ^.) 

(iv.) : when x = p. (Ans. co .) 

(*-PY 



§ 56. Equation to Tangent to a Curve. 

It is clear that, given any equation to a curve, and any point 
on it, we can at once write down the equation to the tangent 
and normal at that point; for we know that the equation 
to any line passing through a point (ab) whose tangent 
of slope is ;;/, is (y — b) = ?;i(x — a); for this equation, being 



Miscellaneous Applications of Differentiation. 157 

of the first degree, clearly represents a straight line. It is 

also satisfied by the point (a, £,), and since it is of the form 

y = mx + c when simplified, it represents a line inclined at 

dy 
tan -1 ;//. Hence, substituting — for m, we have — 

(y — b) = -j-(x-ci) for the tangent 

and (v — b) = — — for the normal 
dy 

dx 

Example 1. — Find the equation to a tangent and normal of 

x 2 
the curve jy = — , at a point on it whose abscissa is 5. 

Here it is clear y = 2*5 ; — = 1. Hence equation to 

tangent is (y— 2*5) - (^ — 5), and to normal jy— 2*5 = 5— jr. 

2. — Prove the subnormal in the curve y 2 = 2?nx is equal 
to m. 

Take any point (x^) on the curve. Since the point is on 
the curve we have — 

y ± 2 = 2mx x 
hence y 1 = V f 2 7?ix 1 

hence the point (x ±i V f 2?nx^) is on the curve. 

Find where the normal at the point (x ly V f 2mx 19 ) cuts OX by 
putting y = in its equation, and show that this point is at a 
constant distance from the point whose abscissa is x lt 



§ 57. Radius of Curvature. 
Consider any curve APQ (Fig. 49). It is clear that y, 

— , etc., are not the only quantities in connection with the 
dx 

curve which assume definite values for any assumed value 



: 5 8 



Graphical Calculus. 



of x. Other such quantities are S, the length of the curve 

reckoned from A ; <j>, the angle of slope in radians, etc. 

We are, therefore, quite within our rights in speaking of 

dx d<j> 
such differential coefficients as — , — (see § 45). It would 

be easy to plot, for instance, a curve showing in its ordinate 

the length of the curve reckoned from A corresponding to 

each value of x. This should be done by measurement. 

The first derived of this curve would represent the value 

ds 
of — . But we can see that another geometrical function 

dx 





Fig. 50. 



Fig. 51. 



would represent this value independently of such a curve. 

PQ 
It is clearly the limit of — -, or the secant of the slope at P, 

which function will vary when the slope of the curve varies. 

ds 
Consider what is the geometrical meaning of — . It is 

d<p 

the limit of the ratio of the length PQ (Fig. 50) to the dif- 
ference (measured in radians) between the angle of slope of the 
curve at P and at Q. Consider first what this ratio means 
for the circle APQ (Fig. 51). 
We have by definition — 

(W<£ = ds, where OR = 1 inch 
ds 



i.e. OP = 



d<j> 



Miscellaneous Applications of Differentiation. 159 

This holds good just as well for the circle in Fig. 50, for 

it is clear that the angle d<j> between PT and QS = the angle 

POQ. OP is here clearly the radius of curvature of a circle, 

which most nearly coincides at P with the curve. Hence, to 

find the value of the radius of curvature at any point on the 

curve, we have to direct our attention to finding the value of 

ds 

— at that point; that is, the height of the first derived curve 

of the curve representing the values of <£ (abscissa) and s 
(ordinate). 

This curve can easily be drawn by measurement. For the 
co-ordinates of any point corresponding to P, Fig. 50, we must 
take AP (the arc) for ordinate, and TK (arc) for abscissa 
where PT = 1 inch. Differentiating this curve graphically, we 
obtain a curve showing in its ordinate the length of the 
radius of curvature. 

The algebraical process is directed towards obtaining the 

ds 
value of — from the successive derived coefficients of the 
dcj> 

primary curve with respect to x. 

We have — 

//c T 

• - (§23) 

• - (§43) 



dy 
Now, tan <£=--. 
ox 

Differentiating this with respect to x (§§ 43, 44), we 
obtam — 

'(E) 



r 


ds 
= d$ 


I 
~~ d4> 
ds 






[ 




dcjy 

dx 


dx 
' ds 



dtsan. (/> \ dx 

dx dx 



160 Graphical Calculus. 

d<t> d 2 y 

or sec- <t>— = — • 

dx dx- 

d<t> d 2 y _ 
Hence -r- = -ji cos- </> 

Also — = cos </> 

i i sec 8 <£ 

Hence r = = — — = — — 

dd> dx d 2 v o . dy 

Now, we know that sec 2 </> = i + tan 2 9. 

2 



Hence sec <£ = \/ 1 4. ( — ) 

\dx ' 



JjYi 



Hence r = 



dx 
d*y 



dx 2 

which gives the value of r in terms of the corresponding height 
of the first derived and second derived curve. 

This process seems confusing at first. The student should bear in mind 

, . . ds . . dy d 2 y 

that the object is to express — m terms of y- and -7=^. In order to do 

this, we must first separate ds and d<f> by means of § 43. By that section 
we know that — 

d<p __ d(p dx 

ds dx ds 

Now, we can find — - by expressing the relation between <f> and x, 
and differentiating it with respect to x. This relation is — 

tan cb = — - 

dx » 

or^ = tan-'- 
On differentiating either of these, we obtain -7- in terms of -7-, and 



Miscellaneous Applications of Differentiation. i6i 

l ~-^- 2 . We also know that -r — cos (/>. These equations are combined, as 
shown in the above article. 

There are several other expressions for this most important 
function, but the use of them involves ideas beyond the scope 
of the present work. 

Example i. — Find the smallest radius of curvature of the 
curve y - e x . 

Example 2.— Prove that, when a ball is projected obliquely 
upwards, the centrifugal force due to the curvature of the path 
at the highest point just balances the weight of the ball. 



vt — A 







Fig. 52. 



At its highest point the ball is moving horizontally (§ 10) 
with velocity z/-— , suppose. Now, after / seconds the co-ordi- 



sec. 



nates of the position of the ball are x = vt, andjy = \gf 
The equation of the path of the ball, therefore, is — 



} 2 V 2lf 



This is found by eliminating / between x = vt (i.) and 



y - \gf- (ii.) j i.e. substituting - for t in (ii.). 

Hence ■—- = — -,x 
ax v* 

fy = ~* 

dx- v 1 



u 



io2 Graphical Calculus. 

Hence radius of curvature at the point (o, o,) 

(i+o 2 )* v 1 

g ~ S 



i.e. a distance — vertically downwards. 
g 
Let m - mass of ball. 
Centrifugal force on ball due to curvilinear path 

VIV 1 _ 1HV 2, 

r v 2 



- mg - weight of ball 



§ 58. Illustration of Taylor's Theorem. 

It may not be out of place in the present work, without 
going into the proof of what is called " Taylor's theorem " 
(which will be found in any book on the differential calculus), 
to give an illustration of the meaning of that very compre- 
hensive proposition, which will, perhaps, enable the student 
to grasp its meaning better. 

The proposition is as follows : If any curve is represented 
by the equation y = f(x), and if we know the height of the 
curve and all its derived curves corresponding to one value x 
of the variable, then assuming that the function is " continuous," 
i.e. that neither the function nor its derived functions become 
infinite for any of the values of x under consideration, the 
height of the curve at a point whose abscissa is (x + h) is — 

Ax) + &(*)■+ £/"(*) + £- 3 rx + . . . (i.) 

where f'(x\ f"(x) 9 etc., are the heights of the first, second, 
etc. , derived curves at the point whose abscissa is x. 



Miscellaneous Applications of Differentiation, 163 



we 



shall take for illustration will be 



The function 
The curve is y = x\ and the height which we shall calcu- 
late corresponds to an abscissa 
(x + A). 

Assume that the curve (Fig. 53) 
represents the distance travelled 
by a particle, as in Fig. 17. 

Let Op = x, pq = h. . 

Then if OPQ represent the 
curve, y=f(x), it is clear that- 
'/Q represents /(# + A). 

Of course, we could in this 
case arrive at the height gQ by 
cubing (x + A) ; but we can also 
arrive at it by another process, 
which has the advantage that it 
is applicable to all other func- 
tions of x. Differentiate the 
primary function. Then it is clear 
that the area P>V'Q' = MQ. 




Fig. 53. 



^Q = ?M + MQ =/(*) + area PWO' 
=/(*)+ P//N + SP'N + Q'P'S 
=/(*) + ¥'(*) + i& X h tan SP'N + area Q'P'S 
= /(*) + W* + iH &f(x) + area Q'P'S' 



Now, this small area Q'P'S' is called " the remainder after 
three terms of the series." An expression is found for it in 
all books on the calculus. Its actual value in this case is A 3 . 
Working the above formula (i.) out, we find — 

f(x) = X s 

fix) = 3* 2 

/'(*) = 3- 2X 
f\x) = 3.2.1 



164 



Graphical Calculus. 



fix + h) = x* + ZocVi + ^A 2 + /^~ + o 
J x ' J 1.2 1.2.3 

= # 3 + 3^'A + sxh 2 + /* 3 

which agrees with the ordinary formula. 

Again, suppose we are given sin 30 = 0*5, and are required 
to find, say, sin 35 . We have — 

x — 30 = - radians 
6 

70S 73 " 
D 180 

= — radians 



Here/(#) = sin x = J 

/'(#) = cos # = — - 

2 

/"(•*) = - sin # = - \ 

f"{x) - —cos x = 

2 

f"\x) = sin x, etc. 



^3 



/r 



7T 2 . 

- — o X i X 0-5 ■ 



- °' S + 36 * 2 3 6 2 ' 



I ^ , 7T 4 I 

X X — + -74 x 

1.2.3 2 3 6 1.2.3.4 



X i, etc. 



= 0*5 + 0*075— 0*0022 

/.<>. sin 35 = 0-573. . . 

The process being carried to any desired degree of accuracy. 
Again, given log 10 2 = 0*301, required log 10 3. 

Here (x) = /x log x 

= 0-434 log # 
A = 1 



Miscellaneous Applications of Differentiation. 165 

m - \ 

/» = -! 

/""(*) = ~% etc. 

Now, x = 2, /i = 1 ; and log 10 2 = 0*301. 
Hence log, (,+,) = 0-30: + °^ -^ + °^, etc. 

= o*53 6 - °*° 61 = °'47S = lo g 3 

The process being, as before, carried to any desired degree 
of accuracy. 



Examples. 

(1) Given log I = o, find the distance between marks I and 2 on a slide 
rule where the distance between I and 10 is 12*5 cm. (The equation to 
the curve is y — p log x, find the value of/.) 

(2) Find by calculation hyp log 3*25, hyp log 4*21, hyp log 7, given 
log 1 = o. 

(3) Find by calculation cos 4 , cos 66°, sin 72 , etc. 

(4) Find the value of 5 2 * 15 , 3 4 ' 31 , etc., by calculation from the values of 
c2 } 3 4 , etc. (The equations are here y = e (x]og5) , etc.) 



CHAPTER XII. 



MISCELLANEOUS APPLICATIONS OF INTEGRATION. 




Fig. 54- 



§ 59. The Cubature of Solids. 

We have already shown the application of integration to the 
finding of areas. 

It was shown in Chapter I., § 3, that a line may represent 
an area. If a line in one direction represents an area, and 

if a line at right angles to it repre- 
sents a linear distance, then it is 
clear that the area of the rectangle 
formed on these two lines as sides 
will represent a volume. Thus, if 
the number of inches in AB (Fig. 
54) represents the sectional area of 
a prism in square inches, and AD = length of prism, it is 
clear that the number of square inches in ABCD represents 
the number of cubic inches in the substance of the prism. 

If the sectional area of the prism is not constant all along 
the length, but varies from point to point, then if a curve BEC 
be drawn so that the length of the ordinate FE at any point 
F represents the value of the sectional area at that point, then 
it is easy to show by splitting the area up into vertical 
elements, exactly as explained in § 13, that the area of the 
figure BECDA still represents the volume of the irregular 
solid. 

Exercise. — Draw any irregular curve about 10 inches long, 



Miscellaneous Applications of Integration. 167 

and a straight line of similar length. Imagine a solid 
generated by the curve revolving about the line as axis. 
Find graphically the whole volume generated by the method 
of sectional areas. 

Now, the algebraical method of obtaining the volume of a 
solid is the counterpart of this process. It consists in obtain- 
ing the equation to the line of sectional areas BEC (Fig. 54), 
and integrating it with respect to x between the ordinates AB 




Fig. 55. 

and DC. Let it be required to obtain the volume of a cone 
of the dimensions shown in Fig. 55. 

To obtain the equation of the line of sectional areas, con- 
sider what will be the height of the curve at a distance x from 
0. Now, the radius of the circular section of the cone at that 

x 
point will clearly be x tan 6 = -^ X x = -, and the area of 

3 

this circle - 



height of curve /P at that point. In other 



1 68 



Graphical Calculus. 



words, the equation to the curve of sectional areas is y - -x\ 

Hence we require the area of this curve between the limits o 
and 15. This is given by — ■ 



l$TTX~ 7T 

— dx - - 



*5 



x^dx 



7T 15 rv 
- - 

9 °L3. 



_ 1 Jl1 _ 5 X 5 X 57T 
27 1 



= 12571- cub. in. 



7T X 

It is easy to show that the formula - X — gives exactly 

the same result as that derived from the common rule, " ^ of 
volume of cylinder on same base;" for, taking r = radius of 
base, // = height, we have, as a result — 



*(?) dx= 9" X ^' 3 



3' 



The formula for the volume of a sphere of radius a is 
obtained in the same way. 

Taking the origin at the centre of the elevation of the 
sphere, the equation to the curve of sectional areas is clearly 
y - 7T (a 2 — x 2 ). The integral is therefore — 

a a V x z 

ir(d 2 — x 2 )dx = ir I a 2 x 

— a — a [_ 3 

The same result may also be obtained by imagining the 
sphere split up into concentric spherical shells, remembering 
that area of surface of sphere is 4tt/- 2 . 



Miscellaneous Applications of Integration. 169 

Example 1. — Obtain the volume of a cone of height /i, 
the base being an ellipse whose semi-axes are a, b (area of an 
ellipse = irab). Ans. ^ -n-ab/i. 

2. — Obtain the volume of a solid paraboloid generated by 
the revolution of the parabola, y 1 = /\ax, round its axis, 
between the planes x - 5 and x = 9. Ans. 112 na. 



§60. 

By a slight extension of this principle, we can obtain such 
results as the following. 

Find the total mass of a sphere of radius 10 inches, whose 
density varies as the square of the distance from the centre, 
the density (mass per unit volume) at the surface being 0*25 lb. 
per inch 3 . Consider an elementary spherical shell of infinitely 
small thickness dx, and of radius x. The surface of this shell 
is 4ttx\ The volume of it is \i:x'dx. The quantity of matter 
in it is clearly ^7rx 2 pdx, where p is the density or quantity of 
matter per unit volume at a distance x from the centre. Now 

we have io* 2 : x 2 : : . _ : p, since the density varies as the 
m. 3 

square of the distance from the centre. 

0*25 x x 2 lbs. 



Therefore p 



in. 



Hence, substituting this value of p in the above expression, 
we see that required total mass is the result of adding together 

all such small masses as f 4?r^ 4 x — | J dx between the given 

limits ; that is — 



1 o- 



47rx i X 7 j dx = X lx" 

o \ \or J \ ico / o 

= X io° = 20077-lbs. 

500 



I/O Graphical Calculus. 

§ 6 1. Graphical Solution of Differential Equations. 

Problems sometimes arise in which we are given a relation 
subsisting between two or more of the primary or successive 
derived functions of a quantity, and we are required to find 
either the primary or some other function connected with it. 
These problems are very confusing to the beginner, and we 
shall show in what way many of them can be attacked graphically 
by the careful application of the principles already explained. 
For example, a train weighs 50 tons exclusive of the engine. 
The resistance to motion due to mechanical friction alone is 
constant at all velocities, and is of the magnitude of say 8 lbs. 
per ton. The resistance due to other causes (such as that due 
to the atmosphere) varies directly as the 17th power of the 

velocity in — — , being, let us say, = '0025 x v 1 ' 7 . Suppose 

we are also given a curve showing the magnitude of the 
pull in the drawbar as the speed varies, and are required 
to find — 

(i.) The maximum velocity attainable on the level. 

(ii.) The time occupied in attaining it. 

(iii.) Distance travelled in that time. 

A method similar to the following may be used in attacking 

problems of this kind. Suppose AHB (Fig. 56) is the given 

velocity-pull curve, of which both scales must be given (the 

figure is not drawn to scale) where the drawbar pull is plotted 

ft. 
vertically in tons suppose, and velocity horizontally in — . ^ 

We have now to draw on the same base the curve of 
resistances. This resistance consists of two parts — 

{a) Frictional resistance, which is constant whatever the 
velocity. 

(b) Other resistances, which vary as the 17th power of the 
velocity. 

Find the total value of (a) for the whole train. 



Miscellaneous Applications of Integration. 171 

This is — 

8 lbs. 



ton 



X 50 tons = 400 lbs. 



Set this value off at O x D to the same scale as the drawbar 




pull is plotted in, and draw a horizontal line DE. This is the 
curve of factional resistances. 

On DE as base plot a curve DB, whose ordinates show 



172 Graphical Calculus . 

the corresponding values of '0025Z; 1 ' 7 to the given scale. Then 
it is clear that the height of the curve DB above O^L x at any 
point shows the total resistance to the motion of the train 
at a constant velocity represented by the abscissa. 

Now, of the total force in the drawbar pulling the train 
only part is required to overcome the actual constant-velocity 
resistance. The whole force over and above this part is 
employed in increasing the velocity of the train, i.e. in produc- 
ing acceleration. This latter surplus force is clearly given by 
the length of the ordinates between the two curves AB and 
DB. At the point B this surplus vanishes ; the maximum 
velocity is therefore given by O& for here the total force in 
the drawbar is absorbed in overcoming constant-velocity 
resistances, and there is none left to increase that velocity. 
Transfer these lengths of ordinate between the two curves to 
corresponding positions on the base 2 X 2 , thus GH = l 2 h, etc. 
This gives a curve ahb.^ which shows the net force producing 
acceleration at all velocities. 

Now, from this curve we can easily deduce another showing 
the actual value of the acceleration produced. We have from 
Dynamics — ■ 



. . . ft. 
mass in tons X acceleration in ; 



Force in tons = 



g 



where g is the acceleration due to gravity, = 32 — '— about. 



sec.- 
Hence — 

Acceleration in - — '— = force in tons x — 

sec.- 50 

It is obvious that we need not consider the mass of the engine in this 
equation, because any force necessary to accelerate the engine does not 
appear in the drawbar at all, being absorbed in increasing the velocity of 
the engine. It is only the surplus force not absorbed in the engine itself 
that appears as a pull in the drawbar. 



Miscellaneous Applications of Integration. 173 

Hence if we reduce all ordinates of curve ahb in the ratio 
f| we shall obtain a curve K 2 L 2 M 2 &>, which gives the actual 

acceleration in — - corresponding to any velocity, 
sec." 

Now, we know that if the velocity and the acceleration were 
each plotted separately on the same time base, the former would 
be the integral of the latter (§ 16), and the problem resolves 
itself into the finding of the time base, i.e. from known simul- 
taneous values of the velocity and acceleration to deduce a 
time velocity and a time acceleration curve. This we may do 
in the following manner : Divide the curve K 2 L 2 M 2 into small 
parts K 2 L 2 , L 2 M 2 , etc., such that each part is nearly straight, 
and draw ordinates at all these points. Take a base O^ as 
shown collinear with 2 X 2 . Let the time from the instant of 
starting be reckoned from 0\ It is obvious that the point 
O y is on the curve of velocities. The height of the acceleration 
curve at this point is clearly 2 K 2 = 0"N, which lines also 
represent the tangent of slope of the velocity curve. Take 
O v S to represent 1 second, and set up ST vertical. Transfer 
the ordinate 2 K 2 to this line as shown ; then, as in § 14, O^K 
must be tangential to the curve of velocities at the point 0\ 
Next consider the point L 2 . Here the acceleration is / 2 L 2 and 
the velocity 2 / 2 . Hence wherever the ordinate of the time- 
velocity curve corresponding to the point L 2 may be, the slope 
of that curve at the point where that ordinate cuts it, must be 
the slope of the line O^L (where L is projected from L 2 ). Hence, 
as shown in § 14, the point on the time-velocity curve corre- 
sponding to L 2 must lie on the line bisecting KO^L, and since 
the height of the point above O r X y is given by 2 / 2 , we can 
find the point P by projecting as shown. This gives another 
point on the velocity curve, and we can proceed exactly as 
before to find a third point. Thus the whole curve may be 
drawn and the acceleration curve plotted as we go along. 
The distance from O" of the point where the acceleration 
curve cuts O n X", gives the required time. The space passed 



174 Graphical Calculus. 

over in this time can be found by integrating the time-velocity 
curve in the usual manner. 

Similar problems can often be solved algebraically if we 
can find the equations to the curves involved, as in the follow- 
ing example : — 

A smooth tube 15 feet long is fixed horizontally to a 
vertical axis, and revolves round it at two revolutions per 
second. A smooth marble is placed in the tube at the vertical 
axis of rotation. Find its velocity when it is swirled out at the 
other end. 

It is clear that if a piece of string of length x were tied to 
the marble, the radial acceleration would be a> 2 #, and the 
tension in the string mo) 2 x, so that when the marble is free 
in the tube, at a distance x from the axis it has an accelera- 
tion ix?x along the tube. Now, if we plot values of a> 2 x along 
a line representing the tube we obtain a curve of accelerations, 
but not a curve of time-acceleration^ so that the area of this 
curve does not represent the velocity, as we see to be the case 
from §§ 14-17 together. If, however, we could by any means 
transform this curve, as in § 43, to a time acceleration curve 
by transferring the ordinates to a time base, then we could 
integrate it graphically. 

We are, in fact, required to integrate is?x with respect to /, 
the time, in order to find the velocity. Hence the problem 
is to find some function which, when differentiated with respect 
to /, will produce a?x. 

Now, we have that — 

fdx\ 
\dt) 



d 2 x 



dx 
that is, f io 2 xdt = ~- 

Consider what is the relation of the element of area u?xdt 
to the element of area u?xdx, which would be the correspond- 
ing element on an x base, i.e. on the length of the tube. 



Miscellaneous Applications of Integration. 175 
It is clear from § 51 that — 

(D 2 xdx = (D 2 X —dt 

at 
and therefore f vPxdx = f u> 2 x — dt . . . (ii.) 

dx 
Now, if we multiply each side of (i.) above by — , we shall 

dt 

find that we are able to integrate both sides with respect to t, 

dx 
and thereby obtain an equation for — . 

„,, r 9 dx _ fd 2 x dx ,' 

Thus f^-.dt^^.-.dl 

[d 2 x dx 
which we see, from (ii.), becomes J i^xdx = I — dt. 

d x dx 
It is easy to see that the expression -z— • —- is the differential 

dt^ dt 

coefficient with respect to / of \ f — J , for, as in § 50, we 

d~x dx 
have split the expression — % — into two factors, one of which 



df 'dt 

' dx 
dt 
Therefore we have — 



/d 2 x\ 

( — ) is the d.c. of the other 



W^ = i\ 



x (dxy 

dt J 



There is no constant required, since, as explained in § 22, 
the line of velocities cuts OX when x = o. 

Hence — = wx 
dt 

The velocity is therefore — 

4Xtj-Xi5 = 6o7t feet per second 



176 Grapliical Calculus. 

Dividing by x } we obtain — 
1 dx 

— — = 0) 

x ' dt 
Integrating again with respect to t — 

log X = 0)/ 

or x - e m -J- const. 

which constant in this case is — 1, as may be seen by con- 
sidering the instant of starting. 

If this latter equation be differentiated twice with respect 
to /, the original equations (i.) will be reproduced. 

A bird's-eye view of the whole of this problem may be 

d~x 
obtained by considering that — = urx, i.e. the height of the 

second derived of the time-distance curve is a positive con- 
stant multiple of the height of the time-distance curve itself. 
The only curve that satisfies this condition is x = e c \ where c is 
a constant. 

Examples. — (1) Given —— = \ ( '— J 2 , find y in terms 
dx 2 * \dx/ 

of x, (i.) graphically and (ii.) analytically. (For the latter, 
divide both sides of the equation by — and integrate. Obtain 



resulting equation for — in the <?form. Invert it, and integrate 



dx 

with respect to y.) Ans. x — —4^ */. 
(2) Solve the tube problem graphically, 



§ 62. Rectification of Curves. 

It is sometimes very useful to be able to find the length 
of curves. 

We saw in the last chapter, § 57, that we might have a 
curve plotted on the x base showing in its ordinate the length 



Miscellaneous Applications of Integration. 177 

of the curve measured from a fixed point on it. The tangent 
of slope of this curve will clearly be the secant of the angle of 
slope of the original curve. 

The student will understand this without difficulty if he 
works the following exercise. Draw a curve (a) of any shape, 
and take a point P on it ; make another curve (b) on an x 
base, by measurement from A, whose ordinates represent the 
corresponding length of curve (a) measured from P. Differen- 
tiate it graphically, and show that the derived curve so obtained 
is the same as would have been obtained by plotting values of 
the secant of the angle of slope of (a) on an x base. Thus it 
is clear that the integrated curve of a curve showing the values 
of the secant of the angle of slope of the original curve is a 
curve showing the length of the original curve. 

Now, if <£ be the angle of slope of the original curve, we 
have, from trigonometry — 

sec 2 tj> - 1 + tan 2 <f>. 

that is, sec <£ = \/ 1 -f ( -^ 

\dx 

/7c 

Sec 6 we have called -— . Hence — 
ax 



length of curve = S = f V J + ( y ' ) 



2 
dx 



the limits being taken as required* 

Thus, find the length of wire rope required to hang between 

two pillars 120 yards apart, assuming the curve of the rope is 

given by the equation — 

mi * * \ 

y = -Kp^e--) 

2 
This is the actual curve in which a rope hangs, and is 
called the " catenary curve." The axis of x is a horizontal 
line at a depth m below the lowest point of the curve ; m also 
represents the length of the same kind of rope which weighs as 
much as the tension at the lowest point of the rope* 

N 



178 Graphical Calculus. 

Assume ;// = 100. 

The sag of the rope is then - y — 100, where/ - greatest 
ordinate. 

Sag of rope = 50^ -f e ~m) — 100 

= 50(1-82 + ^4-) -100 
= 50 x 0-37 = 18-5 yards 

*c = \ l+ K7x) t = *(- + '"-) 

Therefore the whole length, being twice the half-length, 
is — 

since s is measured from the vertex, 

= 100(1-82--^) 

= 127 yards nearly 

Example. — Plot a catenary from the given equation. 1 

Draw a curve on an x base representing the secant of its angle 

of slope, and obtain the above result by integrating this 

curve. 

i 

X 

1 This may best be done by first obtaining the curve y = e m (i.) Dy the 
method described on pp. 102, 103, i.e. find geometrically the lengths of a 
series of equidistant ordinates in geometrical progression, i.e. with a con- 
stant common ratio which must be calculated from the equation. The 

X 

curve y = e m (ii.) can then be found by combining this curve with the 
curve y = — by the method of Figs. 40, 43. These two curves (i.) and 

(ii.) should then be added together (Fig. 26), and the result divided 

. m 
by-. 



Miscellaneous Applications of Integration. 179 



§ 63. Centres of Gravity. 

The finding of centres of gravity and moments of inertia 
is an application of the integral calculus of very great service 
to engineers. The principle of these methods is exactly the 
same as that adopted in elementary mechanics, but the proofs 
are very much simplified by the application of the calculus. 
The proposition relied on in all these methods is that the 
resultant of a system of forces acting on a body has the same 
tendency to twist it about any arbitrary point or line as the 
sum of the twisting tendencies of each of the forces taken 
separately. This proposition, as applied to the special purpose 
before us, is embodied in the following rule. Find the mass 
of each of the separate parts of the object, and multiply each 
by the algebraical distance (+ or — ) of the centre of gravity of 
that part from any convenient straight line. Add all these 
products together, and divide the sum by the sum of the masses. 
The result is the distance of the centre of gravity from the line. 1 

The following graphical process embodies this rule, and 
applies the principles already explained. 

Let it be required to find the centre of gravity of a piece 
of plate cut into the shape of a curve of sines between the 
ordinates x = o and x = 2 (Fig. 58). It will be seen that, for 
the graphical method, any arbitrary curve of any shape what- 
ever might be used ; but as we shall also give the correspond- 
ing analytical process, it will be convenient to consider a 
cun ? of which the equation is known. 

Draw a number of ordinates to the curve at convenient 
and well-defined distances from O, such as 0*2 inch, 0*4 inch, 
o*6 inch, etc. Measure the length of each ordinate, such as 
rR, with a decimal scale, and multiply it by the scale length of 
the corresponding abscissa Or. 

1 A very lucid explanation of this and similar propositions will be 
found in Professor Goodman's treatise on "Applied Mechanics." 



i8o 



Graphical Ca leu h is. 



There is no difficulty in devising a method whereby this may be done 
graphically, if desired. Thus to multiply together the lines AB, AC (Fig. 

57), complete the rectangle, and mark 
off DE = i inch, and complete as 
shown ; then AF represents the pro- 
duct required. The actual ordinate, 
however, may be obtained much more 
rapidly and accurately with a slide 
rule, an instrument with which every 
engineer or scientist should be familiar. 




Fig. 57- 



Plot the value thus found 
along the same ordinate as at 
;-Ri and carefully draw a curve, OR^, through all the points 
so obtained. Now, the area of this curve will be the moment 
of the whole area about O in inch units to the same scale as 
the area of ORQ represents the actual area. That is to say, 
suppose OX, OY are both held horizontally, then the tendency 
in inch units which the force exerted by gravity on the plate 




Fig. 58. 



ORQ has to. twist the plate about the line OY (supposed held 
fixed) is represented by the area of OR^C^Y in square inches, 
to the same scale as the area of ORQ represents the weight 
of the plate. 

Thus if the area of ORQ is = (a) sq. inches, and OR^ 
= (b) sq. inches, and weight of the plate = 2 ounces ; then, 



Miscellaneous Applications of Integration. 181 

since (a) sq. inches represents 2 ounces, 1 sq. inch = ~ ounces, 

2 
and the moment of the weight about OY is therefore b X - 

ounce-inch units. For consider an element of area /Pi of 

breadth dx. It is clear that the moment of the element pY 

about OY is /ix/Px dx X Op, where fi represents the mass 

of a square inch of the plate. Now, p? X Op =-/Pi, and pY 1 

X dx = area of element /P x ; hence the area of the element 

/P 1 x ft = moment of element /P about OY. 

weight of plate 

Now, the quantity //,, as we have seen, = 

area 01 plate 

2 oz 
= - . — - in above example. Hence, to the same scale as the 
a in. 2 

area of /P represents the weight of the corresponding strip 

of plate, pV x represents the moment of that strip about OY. 

The same may be said of all corresponding strips, and is 

therefore true of them all taken together. 

Now, let X be the distance of centre of gravity from OY. 

Then we have area of OR^iQi = area of OPQ x X. 

- area of OR^Q, 

Hence X = — — — 

area OPQ 

Integrate both the curves graphically, as already explained, 
or find their areas by the planimeter or otherwise ; set off a line 
representing the area of the curve OP^ vertically, as at CB 
(Fig. 6) ; set off a line representing the area OPQ to the same 
scale horizontally, as at AC. Join AB, make AM = 1 inch, 
and draw MP vertical ; then MP represents the distance of the 
centre of gravity from OY. 

The process must be repeated with OX vertical to get the 
distance of the centre of gravity from OX. 

We thus get two intersecting lines, each of which contains 
the centre of gravity, which point is therefore found at the 
point of intersection of the lines. 



1 82 Graphical Calculus. 

The student will now have no difficulty in understanding 
the following process, which is inserted without explanation ; 
as it is the same step by step as the graphical process just 
described. 

_ f 2 xsmxdx 

f* o sin xdx 

f [sin x —x cos x] . » \ 
= °— - — — - — - (see § 53) 

o[- cos ^] 

sin 2 — 2 cos 2 — 

— cos 2 + cos o 

sin 114*59° —2 cos 114*59° 

1 —cos 114*59° 

sin 65*41° + 2 cos 65*41° 

1 + cos 65*41 

sin 65° 24' + 2 cos 65° 24 , 

= *—f —5 f — 5 R early 

1 + cos 65 24 

_ 0-90921 + 0*83252 

1*41626 

= 1 "23" nearly 

Examples. — Find the centre of gravity of a triangle, a cone, 
a frustum of a cone, an arc of a circle, a slice of a sphere, a 
rod whose density varies as the ;zth power of its distance from 
one end, any section of a parabolic plate, a theoretical indicator 
card (no compression). 



§ 64. Moments of Inertia. 

Moments of inertia may be found in a similar way to that 
employed for centres of gravity. The moment of inertia of 
an area about an axis in its plane is analogous to what we 
have already described to be the moment of an area about an 
axis; but whereas each ordinate in Fig. 57= ordinate of 
area X distance of ordinate from axis of Y, each ordinate in 



Miscellaneous Applications of Integration. 183 
the corresponding Fig. 59 for the moment of inertia = ordinate 

SA x (O^) 2 . 1 

Let PQRS be any area of which it is desired to find the 
moment of inertia about the axis OY. Set up ordinates 

1 There is a good deal of confusion in the minds of students as to the exact 
connection between the moment of inertia of an area about a line in its plane, 
and what is called the moment of inertia of a solid body imagined spinning 
about an axis. In the discussion on centres of gravity, the same point 
arose in connection with the relation between the geometrical first moment 
of an area about a line and its mechanical analogue. Without going fully 
into a question which has more to do with rigid dynamics than calculus, 
we may point out that a clear conception of the meaning of the moment of 
inertia of a body may be obtained by considering it as the angular mass 
of the body and a couple as angular force. The meaning of this will be 
clear from the following analogy. If a force acts on a mass perfectly free 
to move — 

Force in poundals = mass in pounds X acceleration in feet per second per 
second 

Similarly, if a couple acts on a mass perfectly free to turn round- 
Couple in ft. -poundals = moment of inertia X angular acceleration in 
radians per second per second 
or angular force = angular mass X angular acceleration 
Again — 

Momentum = mass X velocity 
angular momentum = moment of inertia X angular velocity 
, . . [\ mass X (velocity) 2 

klnetlC ener ^ = { II X (angular velocity)' 

A conception of the magnitude of unit moment of inertia in foot and 
pound units may be derived from the consideration that if a body has unit 
moment of inertia round an axis, and is rotating at unit angular velocity, 
it will do \ ft.-poundal of work before being brought to rest. The con- 
nection between the geometrical moment of inertia and the mechanical 
one consists simply in the introduction of the factor p, or the mass per 
unit area, into the expression for the element. The mechanical significance 
of this is easily seen from the above remarks. The geometrical moment 
of inertia is, as it were, a skeleton which we may endow with life either 
by multiplying by pounds mass per square inch ; it then comes in for 
calculating kinetic energies, etc. ; or if used for such purposes as the 
determination of stresses in beams, we multiply it by pounds weight per 
square inch (tension or compression), and in other ways which need not 
be here mentioned. 



1 84 



Graphical Calculus. 



parallel to the given axis, as in § 63, and multiply the length 
SiSa of each by the square of its distance from the given axis 
(Os) 2 , and set up the length so obtained on each of the ordi- 
nates; thus ^S = S^ X OS 2 . The area of the curve so obtained, 
found in any manner, is the moment of inertia of the area 
about that line in inch units. The proof of this is almost 
identical with that given for the similar method used in con- 
nection with finding centres of gravity in § 63, and may easily 
be completed by the student himself. 

Just as before, if y - f(x) be the equation to the boundary 
line of the curve, all we have to do is to find the value of 



/: 



yx 2 dx s which gives us at once the value of the moment of 




Fig. 59. 



inertia. The value of this area or definite integral, divided by 
the greatest distance from the axis of the boundary of the 
figure, gives us what is known as the modulus of the section. 



Miscellaneous Applications of Integration. 185 

It may be found graphically, in the manner explained in § 63 
as the length of a line. Also the value of this area or integral 
divided by the area of the figure gives the value of the square 
of the radius of gyration. 

Thus, required the dynamical moment of inertia of a flat 
circular plate of mass m lbs., whose radius = r, about an axis 
passing through its centre perpendicular to its plane. Con- 
sider a circular element of the plate, radius x, breadth dxj 

Its area is 2-irxdx 

its mass is 2?rxdx X /x = 2irxdx x — 

irr 

where \i is mass of unit area of plate— 

21TlX >> dx 

its moment of inertia about O obviously = - — 




Fig, 60. 

Hence, whole moment of inertia — 

-fydx 



2tn 

7z< 



~(±xT 



r2 

;-2 



The geometrical moment of inertia is -|tf/' 2 or -;- 4 , which, 
multiplying by the mass per unit area, becomes -|;//r 2 as above. 



1 86 Graphical Calculus. 

Consider the case of a cylindrical shaft under torsional 
stress. Suppose f is the shearing stress per square inch at a 
distance = i inch from the axis. Clearly the stress at any 
distance x from the centre is fx. Since the stress varies as 
the strain, and the strain varies as the distance from the axis. 
Area of an elementary annular ring of radius x x breadth dx 

— 2-irx X dx 
total stress on this layer 2-n-xdx xfx- 2irfx h dx 
moment of this stress about axis = 2ir/x 2 dx X x = 2-Trfx'dx 

Plot values of this along the horizontal radius, and find the 
area of the curve so obtained, and compare result with the 
result of the integral of 2irfx'dx between limits r and o. 

Examples. 
i. Find the moment of inertia of a fly-wheel, outside diameter 6 feet, 
sectional area of rim 4X5, inside diameter of rim 5 feet 4 inches, six 
arms each of sectional area an ellipse of axes 35 and 2. Boss, a cylinder 
8 inches diameter X 8 inches long, with a 3-inch hole for the shaft. Mass 
of cubic inch of iron, 0*26 lb. = p. (Method. — Plot a curve on the 
horizontal radius of the wheel as base, showing the value of pax 2 , where a 
is the area of metal cut through by an imaginary cylinder of radius x, 
concentric and coaxal with the wheel.) Find the scale on which the area 
of this curve represents the moment of inertia. Find the radius of gyration 
by a graphical process (find the weight of the wheel graphically as the 
area of a curve, showing the values of pa). Then — 
I = MR 2 

Find this graphically by the process of Fig. 3. 

2. Find the moment of inertia, by graphical method, of a rectangular 
section, a box section, a triangular section, and a circular section about 
axes in their planes and passing through their centres of gravity, and 
compare your results with that given by the formula nah 2 — 
where n = ^ for rectangular section. 
■j E for triangular section, 
^g for circular section. 
a = area of section. 
h — height in plane of bending. 



APPENDIX. 



Barker's Integraph. 

This instrument was devised by the author for the purpose of 
mechanically drawing the integral curves, on the principle explained 
in § 14. It is here described for the first time. It consists of a 
horizontal slide AB, carrying a slider DF, to which is rigidly 
attached the vertical slide CE, which is fitted accurately perpen- 
dicular to AB. The vertical slide CE carries a long slider GH,to 
which is fitted the tracer P, and to which is pivoted at L the rod 
KL, which slides through the piece M. M is itself pivoted on a 
clamp as shown, and the clamp can be secured to any part of the 
piece D, which is graduated. By means of a double parallelogram 
of jointed rods, KL is kept parallel to the piece N, one point of 
which is pivoted on a slider Q, on which a vernier is engraved ; at 
the same time Q is allowed to assume any position on the vertical 
slide. This piece N carries a wheel with points on its periphery, 
and the bearings of the wheel are so attached to the piece N that 
its plane is always parallel to the axis of KL. It is clear that when 
the pointer P traces out any curve, the wheel will roll out its 
integral, for the tangent of angle of slope of the upper curve is 
clearly proportional to the ordinate of the curve traced out by P. 
The adjustments required are, in the case of a curve on a base, that 
the instrument must be so placed that AB is parallel to OX, the 
given base ; also KL must be parallel to AB when the tracer P is 
on the base OX. However, it may be used for finding areas inde- 
pendently of this adjustment, for if the pointer P be placed on the 
curve whose area is required — such as an indicator diagram — and 
the reading of Q taken, and the tracer be then carried round the 
curve to the siarting-point, and Q read again, the difference of the 



1 88 



Graphical Calculus. 



readings gives the area required, in units which depend on the 
position of M on D. This may be adjusted to read in any units 
within the working limits of the instruments. 

The instrument may also be used for differentiating a curve, by 
tracing out a curve with the wheel ; but it is rather difficult to hold 
the wheel sufficiently steady. 




Fig. fix. 



In this form the instrument was found in practice to be of 
somewhat limited range, and in addition, the friction of the several 
sliding pairs was found too great for satisfactory working. To 
remedy these defects the larger form of the instrument was 




Fig. 62. 



[To face p. it 



Appendix. 1 89 

designed and constructed. The plate is a photograph of the two 
forms. 

The principle on which the larger one works is identical with 
what has already been described, that is to say, it consists 
essentially of a wheel rolling on the paper in such a manner that 
at any point its plane is inclined to the horizontal, at an angle the 
tangent of which is always proportional to the corresponding 
ordinate of the curve which it is desired to integrate. The 
movement of this rolling wheel actuates a pencil, which draws 
the required integral curve. 

The instrument consists of a large frame NG mounted on 
wheels HHO in such a manner that it can freely roll in a hori- 
zontal direction parallel to itself. On the frame run two jockey 
carriages, C, D, free to roll vertically on the frame in straight 
grooves. 

The upper jockey D carries the integrating wheel and an 
adjustable pencil Q, and the lower one an adjustable pointer P, 
which is for the purpose of tracing out the curve which it is desired 
to integrate. 

The mechanism, which will be understood from a careful 
inspection of the photograph, is such that, when the pointer is at 
a distance x above or below the zero position, the integrating wheel 
is inclined at an angle whose tangent is proportional to x. 

Suppose the pointer is at zero, that is to say, is on the base 
line OX of the curve, which it is desired to integrate. The tiller 
bar, guided by the wheel B which rolls on the bar E, then compels 
the plane of the rolling wheel to be parallel to the horizontal base, 
OX. If the frame is now moved to the right, the roiling wheel 
rolls on the paper in a direction parallel to the base line, and does 
not, therefore, move the upper jockey on the frame. But when 
the pointer is at a distance x above the base line, the tiller bar 
inclines the integrating wheel at an angle 6 such that tan oc x. 

If the frame is now rolled on the paper the integrating wheel 
rolls in an inclined direction, the tangent of the inclination of 
which is proportional to x, the distance of the pointer above its 
zero position. The rolling wheel carries the upper jockey with it, 
and the vertical movement of the upper jockey, which is read off 
by means of verniers L, is a measure of the area of the lower 
curve above the base line to a scale which is determined by the 
point of attachment of the tangent bar KS. 



190 Grapliical Calculus. 

The pencil, being attached to the upper jockey, draws a curve 
on the paper which is exactly similar to the curve rolled out by the 
integrating wheel. The scale of the upper curve is determined by 
the position in which the clamps R, S, are fixed to the graduated 
horizontal members N of the frame. 



Vifl x 



- *0 



NOTES. 



(i) Page 13, line 15. To be strictly accurate trie 100 feet should be 
measured along the sloping railway, and not horizontally. When the slope 
is no greater than the gradient of an ordinary railway the difference be- 
tween the two measurements is, for practical purposes, inappreciable. 

(2) Page 14, line 15. The word "rate" derived from " ratio" does 
not necessarily embody the idea of time, though this is its common use. 
Thus we speak of a "rate of wages" implying so much money per hour or 
per week. On the other hand, it is common to speak of a "piece rate," 
implying so much money per piece, the latter having no reference to time. 

(3) Section 12. There are two meanings commonly attached to the 
word "zero." Thus when, speaking literally, we say that a man has no 
money, we mean that he is absolutely without any single coin. On the 
other hand, we may say that the difference between two sovereigns is 
nothing, or zero. We do not imply thereby that they are so absolutely 
equal that no difference in weight could be detected between them, but 
that for practical purposes each is equivalent in value to twenty shillings. 
It is probably impossible to imagine two sovereigns so exactly equal in 
weight that no conceivable method of weighing could reveal a difference 
between them. The mathematical conception of zero is a sort of com- 
promise between these two ideas. It is not literally " nothing." It would 
be impossible to conceive or reason about nothing. On the other hand, it 
is not exactly a small quantity so small as to be unimportant, because 
in pure mathematics there are no such quantities. When they occur (in 
equations, for instance) and are " neglected " the resulting equations or 
values are not mathematical equivalents, but only "approximations." 
They may be more or less close approximations, but they are always 
approximations. Absolute equations, on the other hand, are such that no 
conceivable process of measurement or argument could detect any difference 
whatever between the two sides. Perhaps the best concrete illustration of 
a mathematical zero is the difference between § + i + g + T s + etc., to an 
unlimited number of terms, and unity ; or the difference between 0*9 and 
rooo. This difference would be strictly described in abstract mathematics 
as "zero." The above discussion may seem rather like hair splitting, but 



192 



Graphical Calculus. 



it is, in fact, of great importance to grasp exactly the idea of zero as this 
term is used in modern mathematics, as the whole of the fundamental 
reasoning of the calculus depends on it. 

(4) Fig. II. A correct conception of the relation between these two 
curves is of fundamental and even vital importance. Even at the risk of 
being tedious, therefore, it will be well to reiterate and discuss it at full 
length. The student will find that the time spent in a careful consideration 
of this argument is very well repaid. The conception is somewhat elusive 
and confusing to a beginner, and he is therefore advised to follow the whole 
of the reasoning with the greatest care, because it embodies the fundamental 
argument on which the whole of the integral calculus rests, and until he 
understands it with the utmost clearness he cannot be said to possess a 
sound knowledge of the barest elements of the subject. 

Any rectangular area, horizontally placed, may be regarded as being 
" generated " or swept out by the motion of a vertical straight line at right 
angles to itself. As the " generating line" moves towards the right 
(suppose) the area of the rectangle is evidently increasing at a certain rate, 
measured in square inches per inch increase of the " abscissa " or distance 
moved towards the right. 

We are now about to inquire what is the rate at which the area is 
increasing per inch increase of the abscissa. 

Assume the generating line or "ordinate" to be of constant length/ 
inches. When this line moves, say 2 inches to the right, it clearly sweeps 
out an area (2 X p) square inches. In other words, for every inch it moves 
to the right it sweeps out/ square inches area. This rate of increase of the 
area is the same if we imagine the increment of the abscissa to be J inch 
instead of 2 inches. For the area swept out is \p sq. in. per \ inch moved 
to the right, that is, twice this area for I inch, or, as before, p sq. in. per 
inch moved. The same result will clearly be obtained if we imagine the 
increment to be 0*000001 inch, or any other fraction of an inch, because 
whatever large number we divide the inch by, we have to multiply by the 
same number, in order to obtain the rate of increase per inch increment of 
abscissa. The rate is therefore clearly independent of the absolute magni- 
tude of the increment. 

Now, suppose the ordinate is not of constant length, but increases or 
decreases as it moves along, the bottom end of it always tracing out a 
horizontal straight line. The top end will clearly describe a curve (such as 
the lower curve P V B V in fig. 11). In this case it is still true that if the 
ordinate is at any point of the curve of length/, the rate at which the area 
is increasing at that point is p sq. in. per inch increase of the abscissa (read 
again p. 14). 

This will be understood if a very small increment is assumed, so small that the pro- 
portionate variation of the ordinate is inconsiderable. Suppose, for instance, the 
increment to be o'ooi inch, and suppose at this point that the curve is sloping downwards 



Notes, 193 

at 45 deg. The actual value "of the increase of area so produced is (o*ooi X/) less \ery 
closely half a millionth of a square inch. The increase of abscissa is o'ooi inch, and the 
average rate per inch over this small interval is therefore (0*001/ — 0*0000005) X 1000 = 
(P — 0*0005). It is also clear that the smaller the increment is assumed the smaller will 
the 0*0005 become, and when the increment is assumed infinitely small this figure also 
becomes infinitely small. If the student has got into his mind the conception of a 
mathematical zero (read § 12 and Appendix (3) ), he will see that this quantity is an 
example of such a zero. We are therefore within our rights in describing the number/ 
to be the exact value of the rate in question (i at" the point P x , i.e. over an " infinitely 
small" interval or increment containing the point P. 

The fact that the length of the ordinate changes immediately after the 
point P is passed does not alter the fact that at the point P the rate is p 
sq. in. per inch increase of the abscissa. 

This is the first fact to get clearly into the mind, viz. that at any point 
of a curve the rate in square inches at which the area of the curve is 
increasing per inch increase of the abscissa, is the number of linear inches 
in the ordinate. 

Now consider the length of the ordinate of the lower curve under 
another aspect. By its relation to the corresponding point of the upper 
curve, the ordinate of the lower curve represents the rate at which the 
ordinate of the upper curve is increasing per inch increase of the abscissa. 

Therefore it is clear that the ordinate of the lower curve represents two 
rates. 

(1) The rate of increase of the ordinate of the upper curve per I inch, etc. 

(2) ,, ,, the area of 'the lower curve ,, ,, 

Now, there are two fundamental axioms of the integral calculus, which 
may, perhaps, be called respectively the differentials of two of Euclid's 
axioms. 

Euclid says, " If equals be added to equals the wholes are equal." The 
calculus says, "Equals which have always the same rate of increase are 
always equal." 

The following two illustrations will show the sense in which the latter 
axiom may be applied. If two clerks in an office both start with a salary 
of ^"150 per annum, with an annual increase of ^10 per year each, their 
salaries will be always the same. 

Again. If a man has 30^. a week wage when he is 30 years of age, 
and gets an annual rise of is. per week, his wage will always be "equal" 
to his age (numerically), because each increases at the same rate of one unit 
per annum (the divisor unit). Notice that in this case the units are 
different (one shilling per week and one year), yet the equation holds 
numerically, because the divisor unit is the same in each case. 

Again, Euelid says, "If equals be added to unequals the wholes are 
unequal." 

The calculus is more explicit, and says, " Unequals which have always 
the same rate of increase have a constant difference." 

Thus if a man's wage is 35^. a week when his age is 30, and he has an 

O 



194 Graphical Calcitlus. 

annual rise of is. per week, his wage will always be five units greater than 
his age. 

Apply these axioms to the above problem : — 

Here we have two things : (i) the area of the lower curve, and (2) the 
ordinate of the upper one, which have always the same rate of increase per 
inch increase of the abscissa (the divisor unit). This common rate of in- 
crease is represented numerically by the ordinate of the lower curve. 

Therefore it is clear that (1) if the ordinate of the upper curve is at any 
point equal to the area of the lower one, reckoned from the two axes (read 
the note on "Dimensions of quantities," on p. 5), then any and every 
ordinate of the upper curve is also equal to the corresponding area of the 
lower one. 

(2) If, however, the ordinate of the upper curve is not at any one point 
equal to the corresponding area of the lower one, then, since the rate at 
which the two unequals are respectively increasing is always the same at 
every point, it is clear from the second axiom that the difference is constant 
at all points. 

Thus, suppose the area in square inches between the lines O v Y\ O^X\ 
/ N F, and the curve B X P V (which cuts the line O v Y at some point A N suppose, 
not shown on the diagram) is represented by the line pY + 2\ inches. 
Then, since the rate at which pY is increasing per inch increase of the 
abscissa is at all points the same as the rate at which the corresponding 
area under the curve P X B V is increasing per inch increase of the abscissa, it 
follows that if we take any point B v on that curve, the area A x O x £ x B N A v is 
represented by bB + 2\ inches. 

This being the case, it is easy to see that the area Y s p^b y B\ being the 
difference between A v O v ^B x A v and A x O>T v A v , must be equal to (bF + 2J") 
- (pV + 2j") = bB -pB = CB. 

(5) Page 31, line II. We are here supposing that we wish to calcu- 
late the height of the first derived curve corresponding to any point P on 
the primary, and that the co-ordinates of P are known, say, Op = 2 inches, 
and/P = (2) 2 or 4. 

(6) Page 6j, line 2 from bottom. That is, multiply together the 
arithmetical values of each pair of corresponding ordinates, and plot the 
result on the same base as the new ordinates. 

(7) Page 89, line 9. We are here using the notation dx in the same 
sense as we have hitherto used Ax, but observe that the simultaneous 
existence of dy must be borne in mind, along with the fact that whatever 

dx 

the absolute value of dy or dx the ratio -y- is constant for this value of x. 

y dy 

(8) Page 91. In Fig. 36 Y is on the line O x O, while X v is on the 
line A A. 

(9) Page 99, line 6 from bottom. Where /j. has the value 0*434 . . . 
and is known as the " Modulus." 



Notes. 195 

• (10) Page 104. Add as an exercise on this chapter (which is of great 
practical value, and should never be omitted) the following construction : — 

Draw to a large scale (not less than 10 inches equal the unit) the 
rectangular hyperbola xy = 1 in the following manner : — 

Take axes OX, OY, as usual, and take OA, OB, along OX, OY, 
respectively each equal to one unit. Draw AC vertical, and BD hori- 
zontal. Draw a large number of lines radiating from O, each cutting AC 
in E and BD in F. Then for each radiating line draw a horizontal 
through E and a vertical through F. The point where any two correspond- 
ing lines (such as AC, BD) intersect is a point on the required curve 
(verify this by measurement and multiplication, also prove it geometrically). 

Integrate this curve graphically, taking a point along the new OX, one 
unit from O as starting point. Divide all the ordinates of the integral 
curve by 2*3. Each ordinate so divided is the arithmetical logarithm of 
a number represented by the corresponding abscissa. Compare a series of 
these with the logs taken from an ordinary table. 

(11) Page 123, line 19. Where c and e are some arbitrary constants. 

(12) Page 139, line 5 from the bottom. Thus the length of any 
ordinate / 2 P 2 of the second curve (b) shows the value of the f power of 
2 A> tnat * s to sav > tne i power of the corresponding ordinate P x (since 
the latter is equal by construction to 2 A)- The corresponding value of x 
is in this case clearly ij. Now, if we make a new curve (c) having 
abscissae equal to the abscissae of (a), and ordinates equal to the correspond- 
ing ordinate of (b), it will be a curve showing graphically the relation 

between the value of x (abscissa) and of 1 1 -\ — ) 5 (ordinate). 

(13) Page 153, Examples. The following hints should enable the 
student to solve these examples for himself : — 

Example I. — Assume a distance x from the base of the post. Then we 

20 
have — = tan 9. Calculate from this by trigonometry the value of sin 9. 

Find (by differentiating) for what value of x this is a maximum. Find the 
value of sin 9 when x has this value, and multiply by 160. 

Example 2. — Take x as the diameter of the boiler. Then x 2 multiplied 
by 07854 is the sectional area. Calculate the length so that the boiler may 
hold exactly 200 cubic feet. Then, having found in terms of x, the weight 
of the boiler, differentiate and equate to zero. Thus, find the actual value 
of x, and the length, and calculate the weight. Notice the two solutions 
of the equation. What is the significance of the second solution ? 



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